Consider the $n$-dimension hypercube with vertices at $(\pm x,\pm x,\ldots,\pm x)$. Its volume $V_n(x)$ is $(2x)^n$. Imagine we expand the hypercube a little bit from $x$ to $x + \delta$, each $m$-dimension face of the hypercube will contribute an extra hypervolume $V_m(x) \delta^{n-m}$ to the new hypercube. This means:
$$V_{n}(x+\delta) = \sum_{m=0}^n N_{n,m} V_{m}(x)\delta^{n-m}$$
where $N_{n,m}$ is the number of $m$-dimension "faces" of a $n$-dimension hypercube.
One the other hand, binomial theorem tell us:
$$V_n(x+\delta) = 2^n(x+\delta)^n = 2^n\sum_{m=0}^n \binom{n}{m} x^{m} \delta^{n-m}
=\sum_{m=0}^n 2^{n-m}\binom{n}{m}V_{m}(x)\delta^{n-m}
$$
By comparing the coefficients of powers in $\delta$, we immediately get:
$$N_{n,m} = 2^{n-m} \binom{n}{m}$$
In particular,
$$N_{n,n-2} = 2^2 \binom{n}{n-2} = 2^2 \binom{n}{2} = 2n(n-1)$$
Update
Let $N_{n,m}^c$ be the number of $m$-dimension face in contact to a corner in a $n$-dimension hypercube. If you compare the volume of the hypercube $[0,x]^n$ with that of $[0,x+\delta]^n$ like above, you will obtain:
$$N_{n,m}^c = \binom{n}{m}$$
An alternate combinatorial argument go like this.
Since a $n$-dimension hypercube has $N_{n,0} = 2^n$ vertices.
There are $2^n N_{n,m}^c$ ways of picking a corner and then a $m$-dimension face in contact to that corner. We can arrive the same count by picking the $m$-dimension face first. As a result, we again get:
$$2^n N_{n,m}^c = 2^m N_{n,m} \quad\implies\quad N_{n,m}^{c} = 2^{m-n} N_{n,m} = \binom{n}{m}$$
BTW, you $n$ choose $2$ argument also works. In any event,
$$N_{n,n-2}^c = \binom{n}{n-2} = \binom{n}{2} = \frac{n(n-1)}{2}$$
For tesseract, this is $6 \ne 2 n - 3$. Let's say your tesseract is $\{\;(x,y,z,t) : 0\le x,y,z,t \le 1\;\}$.
The six $2$-dimension faces in contact to the origin are:
$$\begin{cases}
x = y = 0\text{ face} &\to&\{0\} &\times& \{0\} &\times& [0,1] &\times& [0,1]\\
x = z = 0\text{ face} &\to&\{0\} &\times& [0,1] &\times& \{0\} &\times& [0,1]\\
x = t = 0\text{ face} &\to&\{0\} &\times& [0,1] &\times& [0,1] &\times& \{0\}\\
y = z = 0\text{ face} &\to&[0,1] &\times& \{0\} &\times& \{0\} &\times& [0,1]\\
y = t = 0\text{ face} &\to&[0,1] &\times& \{0\} &\times& [0,1] &\times& \{0\}\\
z = t = 0\text{ face} &\to&[0,1] &\times& [0,1] &\times& \{0\} &\times& \{0\}
\end{cases}$$
The $n$-simplex has $\binom{n+1}{k+1}$ faces of dimension $k$.
Then $n$-cube has $2^{n-k}\binom{n}{k}$ faces of dimension $k$.
The $n$-crosspolytope (aka $n$-orthoplex) is dual to the $n$-cube,
so it has $2^{k+1}\binom{n}{k+1}$ faces of dimension $k$ (the number of $(n-k-1)$-faces of the cube.)
Yes, all of these polytopes have only triangles or squares as 2-faces; in fact, all the faces of the cube are lower-dimensional cubes, and all the faces of the simplex or the crosspolytope are lower-dimensional simplices. (In fact, every convex polytope of dimension at least five has at least one 2-face which is a triangle or a quadrilateral.)
So the cells of hypercubes are 3-cubes, and cells of simplices or crosspolytopes are tetrahedra.
Best Answer
Here are a couple hints. I will consider the vertices to be length-$n$ bitstrings.
An edge connects two vertices if the vertices differ by a single bit flip. How many edges are incident to a given vertex? If you multiply this by the number of vertices, what would you do next to end up with the total number of edges?
A facet (i.e. $(n-1)$-dimensional cell) may be chosen by first picking a coordinate, fixing its bit, and then letting all the other bits be arbitrary. How many ways are there to do this?