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.
The moment-of-inertia (MOI) tensor is real (no imaginary terms), symmetric, and positive-definite. Linear algebra tells us that for any (3x3) matrix that has those three properties, there's always a set of three perpendicular axes such that the MOI tensor can be expressed as a diagonal tensor in the basis of those axes. These are called the principal axes (or eigenvectors) of rotation, and the physical meaning behind them is that if you rotate the object around one of those axes, the angular momentum will lie along the axis. So one important thing to realize is that there is nothing fundamentally meaningful about off-diagonal elements; you can always rotate your coordinates to get rid of them. If the object has a symmetry axis, then that will be a principal axis. (Though, having a principal axis does not imply any symmetry of the object.)
On the other hand, what if the body is rotating about an axis that isn't one of the principal axes? This is equivalent to writing your MOI in a basis where the rotation axis is one of the basis vectors, in which case there are off-diagonal elements, which is what your question is asking about. So, off-diagonal elements in your MOI are equivalent to having a rotation axis that is not aligned with any of the principal axes. Again, this only happens when your body is not symmetric about the rotation axis.
And what does this mean physically? For one thing, the angular momentum is not aligned with the angular velocity. For example, imagine your object spinning inside a nicely symmetric little satellite in space. You can see its rotation axis, but if the satellite grabs onto the object, it will absorb the angular momentum, and you'll find the satellite spinning on a different axis.
Alternatively, you can think of the expression relating torque and angular acceleration $\vec{\tau} = I \cdot \vec{\alpha}$. An off-diagonal element in the MOI means that if I apply a torque about a certain axis, the object will accelerate its rotation about a different axis.
Best Answer
The meaning of moment of inertia tensor comes from $$\vec{L}=I\vec{\omega}$$
So for example if you consider an object rotating about the $y$ axis with angular velocity $\omega$, you have
$$\left(\begin{array}{c} L_x\\L_y\\L_z \end{array}\right)=\left(\begin{array}{ccc} I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{array}\right)\left(\begin{array}{c}0\\ \omega\\0\end{array}\right)$$ $$=\left(\begin{array}{c}I_{xy}\\I_{yy}\\I_{zy}\end{array}\right)\omega$$
So $I_{xy}\omega$ is the $x$-component of angular momentum when the object rotates about the $y$ axis with angular velocity $\omega$, and similarly for the meaning of the other components.
In general $I_{ij}\omega$ is the $i$-th component of the angular momentum of the object rotating with angular velocity $\omega$ about the $j$-th axis.
For axes along which the rotation of an object will give angular momentum along the same axes as well, then the off-diagonal elements are zero if these axes are chosen as basis. Because in that case you have, e.g., for rotation about the $x$-axis $$\vec{L}=I_{xx}\omega \hat{x}$$ for rotation about $y$-axis $$\vec{L}=I_{yy}\omega \hat{y}$$ for rotation about $z$-axis $$\vec{L}=I_{zz}\omega \hat{z}$$ and hence in general for $$\vec{\omega}=\omega_x\hat{x}+\omega_y\hat{y}+\omega_z\hat{z}$$ $$\vec{L}=\left(\begin{array}{ccc}I_{xx}&0&0\\0&I_{yy}&0\\0&0&I_{zz}\end{array}\right)\left(\begin{array}{c}\omega_x\\\omega_y\\\omega_z\end{array}\right)$$
$I_{ij}$ is talking about the rotation about the $j$-th axis.