For a non-spherical object, there is a unique direction along which the object is "longest", that is, to have the smallest moment of inertia if rotated about an axis with that direction. The material of the object are as close to that axis as can be, compared to other directions.
There's another direction perpendicular to that about which the moment of inertia is maximum.
Then finally we have an intermediate amount of moment of inertia in a third direction perpendicular to the previous two. I lied; that "intermediate" moment of inertia may be the same as the minimum one or maximum one, in which case you have some freedom to pick an arbitrary angle for one axis, but never mind this detail for present purposes.
A spherical object, of course, has the same moment of inertia about any axis, so is boring. You have freedom to pick axes however you like, but never mind that special case either, since it's not interesting.
For the non-special case, we have the unique directions for minimum, maximum, and intermediate moments of inertia. We could name these directions, the 'principal axes', with letters like, oh maybe: 'X', 'Y', and 'Z' and thus have the tensor
$$
I =
\left(
\begin{matrix}
I_{xx} & 0 & 0\\
0 & I_{yy} & 0\\
0&0 & I_{zz}
\end{matrix}
\right)
$$
These three numbers are physically meaningful, giving a general overall measure of size and mass distribution of the object.
But maybe the object is positioned at some crazy angle with respect to things we care about, like our nice level tabletop, our local notion of 'east' and 'north'. So we must rotate the object and its various physical vectors and tensors (and spinors if it's a fermion). An arbitrary rotation is described by three angles (e.g. Euler angles). The fully general $I$ tensor then has six independent quantities. We see nine components, but they count as six due to always being a symmetric tensor.
The physical significance of the off-diagonal components is that you're using a coordinate system not aligned with the principal directions of the object. They tell us nothing interesting about the object itself.
We will work in units where the mass of each face is $1$ and where the length of the side of the cube is $1$.
The contribution to the moment of inertia of each of the top and bottom faces is, using your result for the moment of inertia of a rectangle, $\frac{1}{6}$.
By symmetry, each of the four other faces has the same contribution to the moment of inertia. Let's calculate the contribution of one of them. Let's project the system along the axis about which we are calcuating the moment of inertia. This operation has no effect on the moment of inertia. Our face now becomes a rod. We know the moment of inertia through the center of the rod is $\frac{1}{12}$, but we are calculating the moment of inertia at a distance $\frac{1}{2}$ away from the center of the rod. Using the parallel axis theorem, we get that the moment of inertia is $\frac{1}{12} + (\frac{1}{2})^2 = \frac{4}{12} = \frac{1}{3}$. Then our total moment of inertia is $\frac{1}{6} + \frac{1}{6} + 4*\frac{1}{3}= \frac{5}{3}$. (The first two terms come from the top and bottom face, the last term comes from the four side faces.)
Putting units back in we get the moment of inertia is $I=\frac{5}{3} M a^2$, where $M$ is the mass of a face and $a$ is the side length of the cube (also the side length of a face).
Best Answer
Cross terms appear when the coordinate axes do not pass through the center of mass. That is if you start with a diagonal inertia matrix at the center of mass, when applying the parallel axis theorem cross terms will appear.
In vector form the parallel axis theorem is
$$ {\bf I} = {\bf I}_{cm} - m [{\bf r}\times] [{\bf r}\times] $$
where $[{\bf r}\times] = \begin{pmatrix}x\\y\\z\end{pmatrix} \times = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}$ is the cross product matrix operator.
So if we start with a diagonal inertia at the center of mass, when moved to a different point $(x,y,z)$ the inertia matrix is
$$ {\bf I} = \begin{bmatrix} I_x+m(y^2+z^2) & - m x y & -m x z \\ - m x y & I_y + m (x^2+z^2) & - m y z \\ - m x z & - m y z & I_z + m (x^2+y^2) \end{bmatrix} $$
So if any two of $x$ $y$ or $z$ are zero the result is still a diagonal matrix. This happens when one of the coordinate axis passes through the center of mass. In your case if the $x$ axis goes from the corner towards the center of mass (across diagonal) then $y=0$ and $z=0$ and the criteria is met.
So the question boils down to under which conditions the inertial matrix is diagonal at the center of mass. The answer to this has to do with symmetries. For example, when a particle at a positive $z$ coordinate is matched with an particle at a negative $z$ coordinate then when the cross terms get added up the result is zero. This can be seen from the structure of the inertia matrix (as shown above) if you think of this as a parallel axis theorem for a lot of small little point masses added together.