Fill the $(n-1)\times(n-1)$ board arbitrary with the black and white.now you should just set the parity with the last row and column.
the last cell($a_{n,n}$) will be same color for both last row and last column because of the parity of the $n-1$ first row is equal to parity of the $n-1$ first column and that is because your board is $n\times n$ and $n\equiv n \bmod{2}$ and if rows parity differ from column parity it is contradiction.
Finally the total answer is $2^{(n-1)\times (n-1)} $.
Update I asked generalized version of this question before.
Let $V$ be the vector space over the field with two elements, $\Bbb F_2$, generated by all matrices of shape $N\times N$, $N=18$, which have exactly one row, or exactly one column with one entries, all other entries being zero. We consider the following map from $V$ to $\Bbb F_2$:
$X$ in $V$ is mapped to
$$
f(X)=
(x_{11}+x_{22}+\dots+x_{N-1,N-1}+x_{N,N})
+
(x_{12}+x_{23}+\dots+x_{N-1,N}+x_{N,1})\ .
$$
(Sum on two consecutive fixed diagonals of $X$, where the indices are considered modulo $N$.)
Then each generator of $V$ is mapped to zero, since the "bits" taken in $f(X)$ are exactly two in each row and/or column. So $f$ vanishes on $V$. But it does not vanish on any quadratic submatrix $A=A(I)$ with ones exactly on the positions $I\times I$, $I$ being a proper interval of $\{1,2,\dots,N\}$. (For instance $I=\{1,\dots,16\}$ for our $N=18$.)
So in our case the answer to the problem is negative.
(The "change of the color on a line/columns" corresponds to adding a generator of $V$ to a matrix. We are starting with a zero matrix. The matrices that can be obtained are exactly those in the vector space $V$ of dimension $N+N-1$.)
Note: I can provide some simple computer check in sage, if needed.
Later edit: It is simpler to test the equations
$$
x_{is}-x_{js}-x_{it}+x_{jt}=0\ ,\qquad 1\le i<j\le N\ ,\ 1\le s<t\le N\ ,
$$
that can be extracted from all $2\times 2$ minors of a matrix $X$ in order to test/decide if a matrix $X$ can be realized. (The minus is also a plus, but makes it simpler to see that e.g. each of the above "one row matrix of ones" satisfy the equation, explicitly $1-1-0+0=0$, and correspondingly for the column matrices in the base of $V$...)
Best Answer
Here is a solution for $n=12$, that you can adapt for any (odd or even) $n>4$. Draw a checkerboard, but make sure the four inner corners stay white.