05AB1E, 3 bytes

7ÝΛ

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Explanation

      # implicit input as length

      # implicit input as string to print

7Ý    # range [0...7] as directions

  Λ   # canvas print

See this answer to understand the 05AB1E canvas.

JavaScript (ES6), 114 106 105 104 103 bytes

n=>(g=x=>v=x*2>w?w-x:x,F=x=>~y?`# 

`[~x?(h=g(x--))*g(y)>0&h+v!=n|n>h+v:(y--,x=w,2)]+F(x):'')(y=w=--n*3)

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How?

This builds the output character by character.

Given the input $n$, we compute:

$$n'=n-1\w=3n'$$

For each character at $(x,y)$, we compute $(h,v)$:

$$h=w/2-left|x-w/2right|\v=w/2-left|y-w/2right|$$

The cells belonging to the octagon satisfy one of the following conditions:

• ($h=0$ OR $v=0$) AND $h+vge n'$ (in red below)


• 
  $h+v=n'$ (in orange below)

For example, with $n=4$ (and $n'=3$):

$$begin{matrix}(0,0)&(1,0)&(2,0)&color{red}{(3,0)}&color{red}{(4,0)}&color{red}{(4,0)}&color{red}{(3,0)}&(2,0)&(1,0)&(0,0)\
(0,1)&(1,1)&color{orange}{(2,1)}&(3,1)&(4,1)&(4,1)&(3,1)&color{orange}{(2,1)}&(1,1)&(0,1)\
(0,2)&color{orange}{(1,2)}&(2,2)&(3,2)&(4,2)&(4,2)&(3,2)&(2,2)&color{orange}{(1,2)}&(0,2)\
color{red}{(0,3)}&(1,3)&(2,3)&(3,3)&(4,3)&(4,3)&(3,3)&(2,3)&(1,3)&color{red}{(0,3)}\
color{red}{(0,4)}&(1,4)&(2,4)&(3,4)&(4,4)&(4,4)&(3,4)&(2,4)&(1,4)&color{red}{(0,4)}\
color{red}{(0,4)}&(1,4)&(2,4)&(3,4)&(4,4)&(4,4)&(3,4)&(2,4)&(1,4)&color{red}{(0,4)}\
color{red}{(0,3)}&(1,3)&(2,3)&(3,3)&(4,3)&(4,3)&(3,3)&(2,3)&(1,3)&color{red}{(0,3)}\
(0,2)&color{orange}{(1,2)}&(2,2)&(3,2)&(4,2)&(4,2)&(3,2)&(2,2)&color{orange}{(1,2)}&(0,2)\
(0,1)&(1,1)&color{orange}{(2,1)}&(3,1)&(4,1)&(4,1)&(3,1)&color{orange}{(2,1)}&(1,1)&(0,1)\
(0,0)&(1,0)&(2,0)&color{red}{(3,0)}&color{red}{(4,0)}&color{red}{(4,0)}&color{red}{(3,0)}&(2,0)&(1,0)&(0,0)end{matrix}
$$

Charcoal, 5 bytes

ＧＨ*Ｎ#

My first answer with Charcoal!

Explanation:

ＧＨ*Ｎ#      //Full program

ＧＨ          //Draw a hollow polygon

   *         //with 8 sides

    Ｎ       //of side length from input

      #      //using '#' character

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Canvas, 15 14 12 bytes

／⁸⇵╷+×＋：⤢ｎ╬┼

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Explanation:

/             a diagonal of length n

 ⁸            the input,

  ⇵           ceiling divided by 2, (storing the remainder)

   ╷          minus one

    #×        repeat "#" that many times

      +       append that to the diagonal

       :⤢n    overlap that with its transpose

          ╬┼  quad-palindromize with the overlap being the remainder stored earlier

Alternative 12-byter.

R, 122 117 115 bytes

function(n){n=n-1

m=matrix(0,y<-3*n+1,y)

v=t(h<-(w=3*n/2)-abs(row(m)-1-w))

m[h*v&h+v-n|h+v<n]=' '

write(m,1,y,,"")}

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Ports the logic from Arnauld's answer, specifically this revision in case there are further improvements. Another 2 bytes saved thanks to Arnauld's suggestion of inverting the logic!

Python 2, 96 bytes

a=b=n=input()

while a>2-n-n:a-=1;b-=a/~-n+1;s=(-~b*' '+'#').ljust(n);print s+s[-1]*(n-2)+s[::-1]

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Python 2, 81 bytes

a=d=n=input()-1

while a<=n:print' '*a+'#'+' #'[a==n]*(3*n-a+~a)+'#';d-=1;a-=d/n+1

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Python 2, 75 bytes

a=d=n=input()-1

while a<=n:print' '*a+`' `'[a==n]*(3*n-a+~a)`;d-=1;a-=d/n+1

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If mixing output characters is OK.

Powershell, 91 bytes

param($n)($s=' '*--$n+'#'*$n+'#')

--$n..0+,0*$n+0..$n|%{' '*$_+"#$(' '*(3*$n-2*$_+2))#"}

$s

PowerShell, 107 97 bytes

param($n)($z=$n-1)..1+,0*$n+1..$z|%{" "*$_+"#"+($x=" "*($z-$_))+(" ","#")[!($_-$z)]*($n-2)+"$x#"}

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If there was a cheap way to reverse the first half, this answer would feel a lot better. It builds the left half, then the core (which is either x #'s or spaces), then mirrors the left's logic to make the right. Fun fact, you don't need to copy over trailing white-space.

Unrolled and explained:

param($n)

($z=$n-1)..1 + ,0*$n + 1..$z |%{  #Range that repeats 0 n times in the middle

" "*$_ + "#" +($x=" "*($z-$_)) +  #Left side

(" ","#")[!($_-$z)]*($n-2) +      #Core that swaps when it's the first or last row

"$x#"}                            #Right side which is left but backwards

C (clang), -DP=printf( -DF=for(i + 179 = 199 bytes

i;*m="%*d%*dn";g(n){P"%*d",n,0);F=0;i<n-1;i++)P"%d",0);P "n");}

f(n){g(n);F=1;i<n;i++)P m,n-i,0,n+i+i-1,0);F=0;i<n-2;i++)P m,1,0,3*n-3,0);F=n-1;i;i--)P m,n-i,0,n+i+i-1,0);g(n);}

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Ungolfed:

f(n){

	int i;

	printf("%*d",n,0);

	for(i=0;i<n-1;i++){

		printf("0");

	}

	printf("n");

	for(i=1;i<n;i++){

		printf("%*d%*dn",n-i,0,n+i+i-1,0);

	}

	for(i=0;i<n-2;i++){

		printf("0%*dn",n+n+n-3,0);

	}

	for(i=n-1;i>0;i--){

		printf("%*d%*dn",n-i,0,n+i+i-1,0);

	}

	printf("%*d",n,0);

	for(i=0;i<n-1;i++){

		printf("0");

	}

}

Python 2, 130 bytes

def f(n):

 a=[' '*~-n+n*'#']

 b=[' '*(n-i-2)+'#'+' '*(n+2*i) +'#'for i in range(n-2)]

 return a+b+['#%*s'%(3*n-3,'#')]*n+b[::-1]+a

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On mobile, so not incredibly golfed.

Batch, 260 bytes

@echo off

set s=

for /l %%i in (1,1,%1)do call set s= %%s%%

echo %s% %s: =#%

call:c %1,-1,3

for /l %%i in (1,1,%1)do echo   #%s:~2%%s%%s:~2%#

call:c 3,1,%1

echo %s% %s: =#%

exit/b

:c

for /l %%i in (%*)do call echo %%s:~,%%i%%#%%s:~%%i%%%s%%%s:~%%i%%#

Outputs two leading spaces on each line. Explanation: Batch has no string repetition operator, limited string slicing capability and requires separate statements to perform arithmetic. It was therefore golfiest to make up a string of the input length in spaces (Batch can at least translate these to #s for the top and bottom lines) and then slice from or to a specific position ranging from 3 to the length to generate the diagonals (this is what the last line of the script achieves).

Ruby, 96 bytes

->n{[*(n-=2).step(z=n*3+2,2),*[z]*n,*z.step(n,-2)].map{|x|([?#]*2*('# '[x<=>n]*x)).center(z+2)}}

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Not very golfed yet. Might golf if I find the time.

Red, 171 bytes

func[n][c:(a: n - 1)* 2 + n

b: collect[loop c[keep pad/left copy"^/"c + 1]]s: 1x1 s/1: n

foreach i[1x0 1 0x1 -1x1 -1x0 -1 0x-1 1x-1][loop a[b/(s/2)/(s/1): #"#"s: s + i]]b]

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Explanation:

Red

f: func [ n ] [

    a: n - 1                                         ; size - 1

    c: a * 2 + n                                     ; total size of widht / height 

    b: collect [                                     ; create a block

        loop c [                                     ; composed of size - 1 rows

            keep pad/left copy "^/" c + 1            ; of empty lines of size c (and a newline)

        ]

    ]

    s: a * 1x0 + 1                                   ; starting coordinate

    foreach i [ 1x0 1 0x1 -1x1 -1x0 -1 0x-1 1x-1 ] [ ; for each offset for the 8 directions

        loop a [                                     ; repeat n - 1 times  

            b/(s/2)/(s/1): #"#"                      ; set the array at current coordinate to "#"

            s: s + i                                 ; next coordinate

        ]        

    ]

    b                                                ; return the block 

]

APL (Dyalog Unicode), 46 bytes^SBCS

(' '@~5 6∊⍨1⊥⊢∘,)⌺3 3⊢<(⍉⌽⌊⊢)⍣2∘(∘.+⍨∘⍳¯2+3×⊢)

This solution was provided by Adám - thanks!

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My (almost) original solution:

APL (Dyalog Unicode), 61 bytes^SBCS

(((⊃∘' #'¨1+5∘=+6∘=)⊢)1⊥⊢∘,)⌺3 3⊢<(((⊖⌊⊢)⌽⌊⊢)(∘.+⍨(⍳¯2+3×⊢)))

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Thanks to Adám for his help!

The idea is to find the "diamond" that lies partly in the square and apply an edge-detect filter to "outline" the octagone.

Perl 5, 201 197 188 187 186 bytes:

$a=<>;$b=3*$a-4;$c='$"x($e-$_)."#".$"x$f."#n"';$e=($b-$a)/2+1;$d=$"x$e."#"x$a.$/;$f=$a;print$d,(map{(eval$c,$f+=2)[0]}1..$a-2),("#".$"x$b."#n")x$a,(map{$f-=2;eval$c}reverse 1..$a-2),$d

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Reads the size of the octagon from first line of STDIN.

C (gcc), 158 153 bytes

• Saved five bytes thanks to ceilingcat.

O,c,t,g;o(n){for(O=2*~-n,t=c=O+n;t--;puts(""))for(g=c;g--;)putchar(32+(!t|t>c-2?g>n-2&g<=O:t<n-1|t>O?t+O==g|t-O==g|c-g-t-n==n-1|c-g-t+n==3-n:!g|g>c-2));}

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Python 3, 224 bytes

n=int(input())

z=" "*(n-1)+"#"*n+" "*(n-1)

print(z)

for i in range(n-2):print(" "*(n-i-2)+"#"+" "*(i*2+n)+"#")

print((("#"+" "*(n*3-4)+"#n")*n)[:-1])

for i in range(n-3,-1,-1):print(" "*(n-i-2)+"#"+" "*(i*2+n)+"#")

print(z)

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