Solution:

Answer: $118$

Divide the $8$ vertices of the cube into two sets $A$ and $B$ such that each set contains $4$ vertices, any two of which are diagonally adjacent across a face of the cube. We do casework based on the number of vertices of each color in set $A$.

- Case 1: $4$ black. Then all the vertices in $B$ must be black, for $1$ possible coloring.

- Case 2: $3$ black, $1$ white. Then there are $4$ ways to assign the white vertex. The vertex in $B$ surrounded by the black vertices must also be black. Meanwhile, the three remaining vertices in $B$ may be any configuration except all black, for a total of $4\left(2^{3}-1\right)=28$ possible colorings.

- Case 3: $2$ black, $2$ white. Then, there are $6$ ways to assign the $2$ white vertices. The $4$ vertices of $B$ cannot all be the same color. Additionally, we cannot have $3$ black vertices of $B$ surround a white vertex of $A$ with the other vertex of $B$ white, and vice-versa, so we have a total of $6\left(2^{4}-2-4\right)=60$ possible colorings.

- Case 4: $1$ black, $3$ white. As in case $2$, there are $28$ possible colorings.

- Case 5: $4$ white. As in case $1$, there is $1$ possible coloring.

So there is a total of $1+28+60+28+1=118$ possible colorings.