How many edges, faces, cells in a $2\times 2 \times 2 \times 2$ hyper cubic lattice? If I have a $2\times 2\times 2\times 2$ hyper cubic lattice, how many corners, edges, faces, and cells will it be composed of? E.g. the 4D analogue of figure below. Assume the faces within the figure are filled in too. This figure has $36$ 2D faces (1x1 each), $27$ vertices, and $54$ 1D edges of length 1. (Some of the faces of the lattice are shared by more than one cell... and I don't want to double-count here).

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We can perform the same counting argument used to count the $k$-cells of an $n$-cube in order to count the $k$-cells of an $\ell_1\times\cdots\times\ell_n$ hyper-rectangle of $n$-cubes. We will do this by counting the number of marked $k$-cells in the figure (a marked $k$-cell is a $k$-cell with one of its corners "marked"). To create a marked $k$-cell, we first pick a vertex, and then pick $k$ of the edges connected to it to determine the $k$-cell the vertex is a corner of, whenever this is possible. The number of marked $k$-cells with given corner vertex $v$ is $\binom{r}{k}$, where $r=\deg(v)$.

We must determine the number of vertices $v$ with vertex degree $\deg(v)=r$ in the figure, for all possible values of $r$. Consider the $\ell_1\times\ell_2\times\ell_3$ case for inspiration. For the purpose of abbreviation, denote $t_i=\ell_i-1$ for $i=1,2,3$. There are


$8$ vertices $v$ with $\deg(v)=3$
$4(t_1+t_2+t_3)$ vertices $v$ with $\deg(v)=4$
$2(t_1t_2+t_2t_3+t_3t_1)$ vertices $v$ with $\deg(v)=5$
$t_1t_2t_3$ vertices $v$ with $\deg(v)=6$.


Consider "faces" of the hyper-rectangle of various dimensions, defined so that for instance a $3$-rectangle has one $3$-face, six $2$-faces, twelve $1$-faces and eight $0$-faces. Each face is a collection of vertices, and the set of all faces is a partition of the vertex set. For $r=n,n+1,\cdots,2n$ a vertex will have degree $r$ if and only if it is in an $(r-n)$-face.

Let $\Lambda_m$ be the set of $m$-faces. For each $m$-subset $I\subset\{1,\cdots,n\}$ let $\Lambda_m(I)$ be the subset of $m$-faces with dimensions $\prod_{i\in I}t_i$. Note that each $\Lambda_m(I)$ has the same number of $m$-faces, by symmetry. There are a total of $\binom{n}{m}$ $m$-subsets of $\{1,\cdots,n\}$, and there are $2^{n-m}\binom{n}{m}$ $m$-faces in the collection $\Lambda_m$, therefore $|\Lambda_m(I)|=2^{n-m}$ for each $I$. And so $2^{2n-r}e_{r-n}(t_1,\cdots,t_n)$ counts the number of vertices $v$ with $\deg(v)=r$. Multiply this by $\binom{r}{k}$ and sum over $r=n,\cdots,2n$ in order to get the number of marked $k$-cells. Since every $k$-cell has $2^k$ possible corners to mark, this overcounts the number of unmarked $k$-cells by a factor of $2^k$. Dividing, we get

$$\# k\textrm{-cells of an }\ell_1\times\cdots\ell_n\textrm{ hyper-rectangle} =\sum_{r=n}^{2n}2^{2n-r-k}e_{r-n}(\ell_1-1,\cdots,\ell_n-1)\binom{r}{k}.$$
Your hypercube is determined by the vertices $(0,0,0,0)$, $(0,0,0,1)$, $(0,0,1,0)$, $(0,0,1,1)$, etc. Thus there are $2^4$ vertices (one for each quadruple).

How many edges? An edge is two points that differ in only one place, such as $(0,\color{red}0,1,0)$ and $(0,\color{red}1,1,0)$. An edge is determined by the three coordinates that don't change. Thus, there are:
$$\dbinom432^3=32\text{ edges};$$
$\binom43$ ways of selecting three edges and $2^3$ ways of giving them each a zero or one.

How many faces? A face is four points that differ in two places, such at $(0,\color{red}0,1,\color{red}0)$, $(0,\color{red}0,1,\color{red}1)$, $(0,\color{red}1,1,\color{red}0)$, and $(0,\color{red}1,1,\color{red}1)$. A face is determined by the two coordinates that don't change. Thus, there are:
$$\binom422^2=24\text{ faces};$$
$\binom42$ ways of selecting two edges and $2^2$ ways of giving them each a zero or one.

Similarly, there are:
$$\dbinom412^1=8\text{ cells}.$$