Perl, 7×7 (42 non-spaces)

for$i( 
1..56 )
{$_= $i
%8? $i%
7? $":7
: $/;1;
 print}

Output:

      7
     7 
    7  
   7   
  7    
 7     
7      

Marbelous - 16x16

....@0..@1....  
@3..0A08..@2  ..
/\--....>1  ..Hp
..@2..\\  =0@3..
ss..//  --\\\\..
@0/\  @1/\Hp..!!
:s  #the-cake-is
  2020#a-pie/lie

Test it here! Spaces as blanks, cylindrical board, and include libraries all must be checked.

Output

              07
            06  
          05    
        04      
      03        
    02          
  01            
01              

Explanation

There are two boards here: the main board (shown below), and the ss board, which takes no inputs and outputs two spaces (0x20) to STDOUT.

A blank cell is equivalent to a .., and anything after a # is a comment.

Every tick, ss outputs two spaces to STDOUT.

The green path is a simple loop that outputs a newline (0x0A) at the end of every 7th tick.

The blue path will output the numbers (Hp prints a marble as two hex digits) present in the output, at the end of every 6th tick.

After we have printed 01 once, the loop ends, and moves down the red path, which duplicates this marble.

One duplicate is printed (the second 01), and the other is sent down the black path, which terminates the board at the !! cell. Because of the location of the Hp used in this last print, the 01 appears before the same tick's two spaces, rather than after, the behavior of every other Hp call.

Python - 11x11

import re
[print(
    re.sub(
"0",   " ",
bin(x)[
2:].zfill(
11)))for x
in[19,15,
1920,116,
15,1,5,3,
3,3, 67]]

Output

      1  11
       1111
1111       
    111 1  
       1111
          1
        1 1
         11
         11
         11
    1    11

It's a pretty messy and boring solution, but I just thought I'd show that...

1. It can be done in Python
2. If you can compress information about your code shorter than your code, then you can pull off something like this :)

This solution takes advantage of the fact that, if you're within a pair of brackets in Python, then you can split your code over several lines and arbitrarily add spaces without getting an IndentationError. Another way of doing something like this is by ending the line with a backslash.

Python - 7x7 (37 non-spaces)

print( 
'%+7s' 
'\n'%1 
*6+'%' 
'-7s'% 
111111 
      )

Output

      1
      1
      1
      1
      1
      1
111111 

Uses Python's old % string formatting operator to do the job: +7 and -7 take care of right/left justification, and the last space to match the closing parenthesis for print in particular. In preparing the format string, we also have

• automatic concatenation of string literals across lines, and
• string repetition by multiplication (giving us multiple replacement fields for the price of one)

JavaScript (9x9)

 i=9;q=""
; for(;++
i< 91;){;
var q=q+(
!(i% 10.0
)?1:" ");
;i%9|| (q
+="\n") }
alert(q) 

Output

1        
 1       
  1      
   1     
    1    
     1   
      1  
       1 
        1

Notes

I made and golfed (to the best of my ability) code for a square with diagonal of any size n:

q="";for(i=***n***;++i<***n^2+n+1***;i%***n***||(q+="\n"))q+=i%***n+1***?"0":1

replacing the ***asdf*** numbers with constants depending on the side length n, for example for n=6:

q="";for(i=6;++i<43;i%6||(q+="\n"))q+=i%7?" ":1

But, even though that code is length 46, I couldn't get the constant space to line up with a space in the diagonal of the code until it was as big as a 9x9, with a wasted line (the 5th one)

Edit: Changed to add alert(). Before:

 i=9;q=""
; while((
++ i)<91)
{q= q+""+
""+( "")+
(!((i %10
))?1:" ")
;i%9||( q
+="\n")} 

CJam, 5x5, 12 non-spaces

Not a winner, but I wanted to add a rather small and sparse submission, since most answers just print a diagonal.

 1 ]
D * S
 * 5
/ N *
 1 ; 

prints

1 1 1
 1 1 
1 1 1
 1 1 
1 1 1

Test it here.

The last two characters of the code, don't do anything, so it actually only has 10 bytes of real code. For a smaller grid, I could even reduce it by another two bytes to 8, but that doesn't fit on 3x3, and this code doesn't work for even grid sizes.

How it works:

1]           "Push [1].";
  D*         "Repeat 13 times.";
    S*       "Riffle with spaces.";
      5/     "Split into runs of five elements.";
        N*   "Join those with line breaks.";
          1; "Push and pop a 1. No-op.";

Python 3, 8x8

There are 50 non-space characters and 14 spaces. The last line has one useless character, but everything else is necessary.

(*_,q,n 
)=p=''' 
       p
'''[2:] 
print(p 
*2+q*7, 
n+p*4+p 
[0:-1]) 

Output:

       p
       p
ppppppp 
       p
       p
       p
       p
       p

CJam, 5X5

SSSS 
1N]4 
*B11 
    S

And the output is

    1
    1
    1
    1
1111 

I was this close to 4X4 solution. <sigh> See my other answer

Try it online here

Ruby, 8x8

 (0...8)
. map(){
|x |$><<
32. chr*
x+[x ]*'
'<<32 .\
chr*(7 -
x)+?\n} 

Output:

0       
 1      
  2     
   3    
    4   
     5  
      6 
       7

Befunge, 9x9

I have no idea why I did this. It took way too long. I have a massive headache now.

8>v_20gv 
v9<^v#<4 
1@,/<>^6 
v1,*$:<p 
->,!-^87 
:^*25<^g 
_88g,^^4 
9vp\v#6< 
        @

Output:

        @
        @
        @
        @
        @
        @
        @
        @
$$$$$$$$ 

Some explanation

The code uses g to read the @ characters from the grid "on the fly" (and also the final space, which is @ / 2), and p to modify the loop to write the last output line.

Every single character on the code is used at some point, either as code or as data (the 9 and the @ on the two last lines).

I basically had to do a lot of workarounds to make the code work. The instruction pointer does multiple intersections during execution, of which some are jumped over. (I couldn't use any instruction there for different directions as they would interfere. There is no NOP.) Elsewhere I either reused the same character or just undid it (see the $: in the middle).

I also did some creative work on the stack:

• When the (inner) character writing loop terminates (all spaces done for this line), the stack has n,0. I then have to decrement n. The obvious solution would be $1-, but I managed to shorten it by using !-.
• When the (outer) line writing loop terminates (all normal lines printed), the stack has a 0. I then organized code changer (20g46p7g46\p) to use that 0, instead of wasting 2 characters on $0.