Solution:
We can focus on only the black cells of the grid, which we need $32$ of. Moreover, the number of black squares in the bottom row and leftmost column uniquely determine the total number of black squares. Suppose that there are $x$ black cells in the bottom row and $y$ black cells in the leftmost column. Then, each of the $x$ rows with black leftmost cell is identical to the bottom row and has $y$ black cells, while the remaining $8-x$ rows are inverted and have $8-y$ black cells, so the total number of black cells is

$$(8-x)(8-y)+x y=32$$

This rearranges as

$$2(x-4)(y-4)=0$$

which tells us we have $32$ black cells exactly when either the bottom row or leftmost column (or both) contains $4$ black cells.
The bottom-left corner is already black. There are $\binom{7}{3}$ ways to choose three more cells in the bottom row or leftmost column to be black, and $2^{7}$ ways to color the remaining cells in the bottom row or leftmost column with no restrictions. Hence, there are $2^{7}\binom{7}{3}$ ways for the bottom row to have $4$ black cells, $2^{7}\binom{7}{3}$ ways for the leftmost column to have $4$ black cells, and $\binom{7}{3}^{2}$ ways for both to occur. The answer is

$$2\left(2^{7}\right)\binom{7}{3}-\binom{7}{3}^{2}=7735$$