This too long for a comment. I'll post my solution later on if you can't solve it but here's how I've done it (in less than 15 lines of code): I first created a second array (a little big bigger [n+2][n+2]) and did n/2 pass:
0 0 0 0 0 0 0 0
0 0 1 0 0 0 1 0
0 1 0 1 1 1 0 0
0 1 0 1 2 2 1 0
0 0 1 2 2 2 1 0
0 0 0 1 2 2 1 0
0 1 1 1 1 1 1 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 1 0 0 0 1 0
0 1 0 1 1 1 0 0
0 1 0 1 2 2 1 0
0 0 1 2 3 2 1 0
0 0 0 1 2 2 1 0
0 1 1 1 1 1 1 0
0 0 0 0 0 0 0 0
Where a non-zero number x means "I'm the center of a rhombus of size x" (I'm expressing the size in relation with the length of the diagonals [which are both equal in your case] of the rhombus). You can find if you have the center of a rhombus of size (k+1) by checking if {top,right,down,left} are all the centers of rhombus of size k.
The advantage of first creating a bigger array is that it really simplifies your logic but I could do it in place, with a more convoluted logic, by modifying the original array or by using a second array of the same size as the input (once again, it's way easier to simply put a safe "fence" of all-zeroes around your input).
If you don't "surround" your array with a fence, you have a lot of additional if/else checks: this would be prone to errors, lead to bigger code and lead to uglier code.