Here's a picture of the three-dimensional cube.
You're right about the vertices. What do you notice about the edges? In particular, can you split them into parallel families? What do you notice about the coordinates of points that are connected by an edge?
What do you notice about the faces? Can you split them into parallel faces? Do their points satisfy any nice equations, here involving $x_1,\,x_2,$ and $x_3$?
There are more combinatorial hints to be had as well; we don't really need vectors to think about cubes (I personally think it's more difficult with vectors and actual coordinates for vertices, but that's just me)
EDIT:
OK, so the big idea, if we'd like to parameterize (and therefore count) faces, is dimension. Roughly, we'll think of it as how many degrees of freedom we have. What does that mean? Here's an example.
Let's pick the top line in the sample $2$-face. Let's call it $(*, 1, 1)$, because its end points have coordinates $x_2 = 1 = x_3$; we get to choose either $0$ or $1$ for $x_1$. We have one degree of freedom, which agrees with lines being one-dimensional.
If we were to count edges of the three-dimensional cube this way, then, we have to:
Pick the coordinate we'll use an $*$ in; we have ${3 \choose 1} = 3$ choices there.
We also have to pick what we'll make our remaining $3 - 1$ coordinates; we have $2^{3 - 1} = 2^2 = 4$ choices here, since for the $3 - 1$ coordinates, we're choosing between $0$ or $1$.
Thus, we have $3 \cdot 4 = 12$ edges of the one dimensional cube. The fancy way to write this would be to say that we have
$${3 \choose 1} \cdot 2^{3 - 1}$$
faces of dimension $1$ in a $3$-dimensional cube.
How about writing down a two-dimensional face? Well, now we have to choose the spots we'll put two asterisks, and then choose the remaining coordinate, for a total of ${3 \choose 2} \cdot 2^{3 - 2} = 3 \cdot 2 = 6$ faces.
For a four-dimensional cube, the same applies: To specify a $k$-dimensional face, we have to choose where we're going to put $k$ asterisks, and then choose the remaining $4 - k$ coordinates.
Alternatively, and just for fun, you can build cubes inductively (the original post I had intended, but rejected in favor of thinking in terms of coordinates and equations). This utilizes a different style of thinking, it's much more recursive.
