We know that # of faces($2$-cubes) in $n$-hypercube = $n(n-1)2^{n-3}$ and number of vertices ($0$-cubes) is $2^n$.
How many # of $2$-cubes are attached to each $0$-cube in hypercube?
In general, how many # of $k$-cubes are attached to each $h$-cube in hypercube where $0\leq h<k\leq n$?
Is there a generating function for this?
First a more general approach which can be used in several similar combinatorical problems: There are $N_2$ 2-cubes, each of which is connected to $M_2$ 0-cubes. On the other hand, there are $N_0$ 0-cubes, each of which is connected to $M_0$ 2-cubes. Our task is to find $M_0$ in terms of $N_2$, $M_2$ and $N_0$.
We calculate the total number of pairs $(x,y)$ for which $x$ is a 0-cube, $y$ is a 2-cube, and $x$ and $y$ are connected. Because each 2-cube is connected to $M_2$ 0-cubes, the total number of such pairs is $$N_2 M_2.$$ On the other hand, because each $0$-cube is connected to $M_0$ $2$-cubes, the total number of such pairs is $$N_0 M_0.$$ These must be equal, and thus $$M_0 = \frac{N_2 M_2}{N_0} = \frac{n(n-1)2^{n-3} \cdot 4}{2^n} = {n \choose 2}.$$
This answer can also be achieved by an argument specific to the $n$-cube: A vertex can be indiced by its coordinates in $\{0,1\}^n$ and therefore it can be treated as a binary string of length $n$. A face is a collection of four vertices such that they have equal values in $n-2$ coordinates and assume all $2^2$ choices in the remaining $2$ coordinates. For each vertex, there are ${n \choose 2}$ possible choices for the two coordinates that differ in a face, and each of the choices result in a distinct face.