Symmetric Binary Chess?

A symmetric matrix would have $A^{T} = A$ . Note that in the transpose matrix, the diagonal elements remain where they are and other elements get 'mirrored' along the diagonal.If we divide the board into 3 parts, the diagonal, the lower half and the upper half, every arrangement of the lower half will have only one arrangement in the upper half such that $A^{T}=A$ . For this reason, there cannot be an odd number of $1'^s$ outside the diagonal (lower + upper half). Each half has $28$ blocks and the diagonal has $8$ blocks. Therefore in total, there would be $\binom{28}{4}\binom{8}{1} + \binom{28}{3}\binom{8}{3} + \binom{28}{2}\binom{8}{5}+\binom{28}{1}\binom{8}{7} = 368648 \text{ matrices}$