This question has a nice clean answer as long as the camera is far enough away that it can be treated as a "view from infinity", so that the map taking a point of $\mathbb{R}^3$ to its image on the film is linear. If you need to deal with cameras which are close to the cube (so that, for example, only one face might be visible) life is much harder; read up on camera matrices.

So, the nice answer. Let the camera point in direction $(u,v,w)$ with side lengths $(x,y,z)$. Let $\pi$ be the orthogonal projection of $\mathbb{R}^3$ onto $\mathbb{R}^2$ with kernel $(u,v,w)$; we want to find the area of $\pi(\mathrm{cube})$. As Ben Barber says, this is a hexagon which can be dissected into three parallelograms. The sides of these parallelograms are the three ways to chose two of the three vectors $\pi(x e_1)$, $\pi(y e_2)$, $\pi(z e_3)$; let $P_{ij}$ be the parallelogram with sides $\pi(x e_i)$ and $\pi(y e_j)$. So we need a formula for $\mathrm{Area}(P_{ij})$ in terms of $(u,v,w)$. We could slog this out with cross products, but there is a slicker way.

Consider two parallelepipeds in $\mathbb{R}^3$: Let $B$ have sides $(u,v,w)$, $x e_1$ and $y e_2$ and let $C$ have sides $(u,v,w)$, $\pi(x e_1)$ and $\pi(y e_2)$. Since $C$ is formed from $B$ by sliding the vectors $x e_1$ and $y e_2$ parallel to the common side $(u,v,w)$, they have the same volume. The volume of $B$ is $\left| \det(x e_1, y e_2, (u,v,w)) \right|= x y|w|$.

Since $(u,v,w)$ is perpendicular to $P_{12}$, the volume of $C$ is $|(u,v,w)| \ \mathrm{Area}(P_{12})$, and of course $|(u,v,w)| = \sqrt{u^2+v^2+w^2}$. Setting $\mathrm{Vol}(B) = \mathrm{Vol}(C)$, we have
$$\mathrm{Area}(P_{12}) = \frac{x y |w|}{\sqrt{u^2+v^2+w^2}}.$$
The area of the image of the cube is
$$ \frac{x y|w|+x z |v|+y z |u|}{\sqrt{u^2+v^2+w^2}}.$$