More fun with binary chess

We will use this solution as a guide.

First we recall that there are $(n-1)!!$ symmetric permutation matrices of size $n\times{n}$ without any ones on the diagonal, for even $n$ .

How many symmetric permutation matrices of size $n\times{n}$ are there with exactly $k$ zeros on the diagonal, where $k$ is even? We can first place the zeros in $n\choose{k}$ ways, putting ones at the remaining diagonal entries. Now we can cross out the rows and columns containing the diagonal ones and fill in a symmetric permutation matrix with no ones on the diagonal. Thus there are ${n\choose{k}}(k-1)!!$ such matrices.

Finally we can convert any of the $k$ zeros on the diagonal to a one. This process amounts to adding a row to another; thus it preserves the determinant and invertibility.

The reasoning here shows that all the matrices we seek are of this form.

Thus the answer is $\sum{k{8\choose{k}}(k-1)!!}$ , for $k=2,4,6,8$ . This sum comes out to be $\boxed{4256}$ .