A square has $2\cdot2$ sides and $2^2$ vertices, but a three-dimensional cube $C$ has $2\cdot3$ faces and $2^3$ vertices; so you can never cut off all vertices at the same time, using a rotated copy of $C$.
You can, however, do the following: When $C=[{-1},1]^3$ then consider for a suitable $\epsilon>0$ the octahedron $O:=\bigl\{(x,y,z)\ \bigm|\ |x|+|y|+|z|\leq 3-\epsilon\bigr\}\ .\qquad(*)$ This will cut off the eight vertices of $C$ allright. In addition, this idea can be generalized to $n$ dimensions, $n\geq2$.
That $(*)$ describes an octahedron is seen as follows: The vertex $(1,1,1)$ in the first octant is cut off in a symmetric way with respect to the three axes by means of the condition $x+y+z\leq 3-\epsilon$. Writing $|x|+|y|+|z|\leq3-\epsilon$ instead of $x+y+z\leq 3-\epsilon$ makes the resulting solid symmetric with respect to all three reflections $x\mapsto -x$, $y\mapsto -y$, $z\mapsto -z$. In $n$ dimensions the corresponding condition would read as $|x_1|+|x_2|+\ldots+|x_n|\leq n-\epsilon$.