For a hyper cube of dimension N what number of its vertices can be covered by intersection with a hyperplane If you look at intersecting a binary cube (the set contained by [0,1]^n) with a plane in $\mathbb{R}^3$. Then the plane can potentially intersect with the corners of the cube. Depending on the choice of plane you might intersect with 0,1,2,3 or 4 corners, but it’s impossible to intersect with any other number of corners.
If we shift our attention to $\mathbb{R}^4$ it gets a little trickier, all the $\mathbb{R}^3$ numbers must be valid but we also know that hyperplanes such as $x_0+x_1+x_2+x_3=2$ let you intersect with 6 corners.
This leads to our problem:
Given a dimension $n$, what does the list of possible intersection counts of a hyperplane with vertices of the Boolean cube look like? Even for $n=4$ at best we can show the list includes (but as of right now I haven’t exhausted) 0,1,2,3,4,6,8.
Powers of 2 and Pascal’s triangle numbers are all present but there definitely are more exotic coverings out there especially for $n=5$ onwards
 A: This is not a complete answer, but was too long to be a comment.
For $n=4$ your list is complete.
First, note that given a hyperplane, we can split the $n$-cube $C_n$ into two disjoint $(n-1)$-cubes along an axis so that the intersection of the plane with each of the $(n-1)$-cubes is a hyperplane in $\mathbb{R}^{n-1}$. In fact, only an axis normal to the chosen hyperplane can fail to work here. So every solution in dimension $n$ is a sum of two solutions in dimension $n-1$. (Note that in this decomposition, the two $(n-2)$-dimensional hyperplanes are parallel within $C_{n-1}$ if projected into the same $(n-1)$-dimensional space by collapsing our axis of the cube.)
This shows that $2^n$ is always an upper bound for the number of intersections $k$ in $C_n$, so it remains to rule out the cases of $k=5$ and $k=7$.
Suppose a hyperplane intersects $C_4$ in $7$ points. Then it must be formed from two parallel planes in $C_3$, one of which intersects in $4$ points and one of which intersects in $3$ points. But all planes passing through $3$ points in $C_3$ have the same orientation, up to rotation, and their translates all pass through $0,1,$ or $3$ points.
Suppose a hyperplane intersects $C_4$ in $5$ points. Then it is formed either from a $3$-point intersection and a $2$-point intersection, or a $4$-point intersection and a $1$-point intersection. The former case is ruled out by the same reasoning as the above paragraph, while the latter case is dealt with by noting that there are two kinds of $4$-point planes in $C_3$: one containing all the points in face, and one "diagonal" cut whose intersection with $C_3$ consists of the vertices of a $1$ by $\sqrt{2}$ rectangle. Translates of these planes intersect $C_3$ in $\{0,4\}$ or $\{0,2,4\}$ points respectively, which rules out the second case.
For $n=5$, the possible intersections (by the sum decomposition proven above) are a subset of $\{{\color{green} 0}, {\color{green} 1}, {\color{green} 2}, {\color{green} 3}, {\color{green} 4}, {\color{green} 5}, {\color{green} 6}, {\color{green} 7}, {\color{green} 8}, \color{red} 9, \color{green}{10}, \color{red}{11}, \color{green}{12}, \color{red}{14},  {\color{green}{16}}\}$, where I have colored green the cases which I can find examples of.
The $7$-point case here is $x_1+x_2+x_3+2x_4+2x_5=3$; everything else is naturally given by a $4$-simplex, an element of Pascal's triangle, or a doubling of the $n=4$ case. I found no other examples after searching through $1,000,000$ random planes with integer coefficients from $0$ to $10$ for each coordinate.
For $n=6$, I have found examples for $k=0,1,2,3,4,5,6,7,8,9,10,11,12,14,15,16,20,24,32$.
If I am correct that the $12$ values of $k$ given above are complete, then the sequence of the number of solutions for $C_n$ which starts $2,3,5,7,12,(19-28),\ldots$ does not seem to have any good matches in the OEIS.
