First we name the elements of the square as follows: $a_{11}, a_{12}, a_{13}, a_{14}, a_{21}, a_{22}, a_{23}, a_{24}, a_{31}, a_{32}, a_{33}, a_{34}, a_{41}, a_{42}, a_{43}, a_{44}$. We claim that for any given values of $a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}, a_{32}$, and $a_{33}$ (the + signs in the diagram below), there is a unique way to assign values to the rest of the entries such that all necessary sums are even. Taking additions mod 2, we have $a_{14}=a_{11}+a_{12}+a_{13}$, $a_{24}=a_{21}+a_{22}+a_{23}$, $a_{44}=a_{11}+a_{22}+a_{33}$, $a_{42}=a_{12}+a_{22}+a_{32}$, $a_{43}=a_{13}+a_{23}+a_{33}$. Since the 4th column, the 4th row, and the 1st column must have entries that sum to 0, we have $a_{34}=a_{14}+a_{24}+a_{44}=a_{12}+a_{13}+a_{21}+a_{23}+a_{33}$, $a_{41}=a_{42}+a_{43}+a_{44}=a_{11}+a_{12}+a_{13}+a_{23}+a_{32}$, $a_{31}=a_{11}+a_{21}+a_{41}=a_{12}+a_{13}+a_{21}+a_{23}+a_{32}$. It is easy to check that the sum of entries in every row, column, and the main diagonal is even. Since there are $2^{8}=256$ ways to assign the values to the initial 8 entries, there are exactly 256 ways to fill the board.