Actually its easy to to figure that recursive relations can help , and here we can easily find out that if a(n) is the answer for a n*n square then a(n) = 4a(n-1).

The answer is $32$. Also, it's more easy if you just look to the chessboard mod 2. For $n=3$, you can check that if you choose a number for $a_{1,1}$, $a_{1,2}$, $a_{1,3}$, $a_{2,1}$ and $a_{2,2}$, you can complete all the chessboard. So, $2^5=32$ ways. Now, you can just use induction and complete a $n-1 \times n-1$ chessboard inside the $n \times n$ and them, it's easy to check that you can complete the remaning cells.