i know you have possibly seen my last problem on the 8 Queens conjecture, and it was answered by a person named Peter Kagey. he mentioned that:
With regard to a $n×n×n$ chessboard, one could simply place $n$ queens at the "bottom board" of the cube, and use the $n*n$ configuration. This argument shows that A000170 would serve as a lower bound on the number of configurations on the $n×n×⋯×n$ analog of the nonattacking queens problem.
If the sequence A000170 is the lower bound, what is the maximum number of queens that can go onto a 3-D chessboard? 4-D? so on?
EDIT I am now just needing the $3\times 3\times 3$, $4\times 4\times 4$, and $5\times 5\times 5$ chess "cubes" i hope it isn't such a complex algorithm or massive time crunch.