The answer is $\frac{n^2-1}{2}$ if $n$ is odd and $\frac{n^2-4}{2}$ if $n$ is even.

Notice that:

* the unit squares from the corners are white (because they only have two neighbors, not three);
* the other squares from the border of the board have only three neighbors, so there can't be two consecutive black squares on the border;
* any $2 \times 2$ square contains at most two black squares (if not, a black square would already have two black neighbors, so it would have at most two white neighbors).

If $n$ is odd, color the board like a chessboard with white corners. This coloring satisfies the requirements of the problem and it has $\frac{n^2-1}{2}$ black squares, so the maximum number of black squares is at least $\frac{n^2-1}{2}$.

On the other hand, split the board in 4 parts: one upper-left $(n-1) \times (n-1)$ square, a unit square in the right bottom corner, a $(n-1) \times 1$ rectangle on the right side and a $1 \times (n-1)$ rectangle on the bottom side. Tiling the $(n-1) \times (n-1)$ square with $2 \times 2$ squares and the two rectangles with 2-square dominoes, according with the remarks made at the beginning, one can have at most
$$\frac{(n-1)^2}{2} + \frac{n-1}{2} + \frac{n-1}{2} + 0 = \frac{n^2-1}{2}$$
black squares.

If $n$ is even, again consider a chessboard-like coloring. This makes two white corners and two black corners; recolor the black corners as white. The configuration we have now satisfies the requirements of the problem and it has $\frac{n^2-4}{2}$ black squares, so the maximum number of black squares is at least $\frac{n^2-4}{2}$.

Split the board in 9 parts: one $(n-2) \times (n-2)$ square in the middle, four unit squares in the corners and four $(n-2) \times 1$ or $1 \times (n-2)$ rectangles on the borders. Tiling the $(n-2) \times (n-2)$ square with $2 \times 2$ squares and the rectangles with 2-square dominoes, the remarks made at the beginning lead to a maximum of
$$\frac{(n-2)^2}{2} + 4 \cdot \frac{n-2}{2} + 4 \cdot 0 = \frac{n^2-4}{2}$$
black squares. Hence, if $n$ is even, the maximum number of black squares is $\frac{n^2-4}{2}$.