This is sometimes called the "tennis racket theorem". It can be derived from a perturbative solution to Euler's equations (torque free).
If you have rotation around any of the 3 principle axes, perturbations about 2 of the axes will lead to nice oscillations in the rotation (I forget if it's called nutation or w/e), but perturbations about the intermediate axis will lead to exponential growth in the perturbation (and therefore violating the validity of your perturbative solution).
Euler's equations (torque free):
I_1\frac{d}{dt}\omega_1=(I_2-I_3)\omega_2\omega_3
I_2\frac{d}{dt}\omega_2=(I_3-I_1)\omega_3\omega_1
I_3\frac{d}{dt}\omega_3=(I_1-I_2)\omega_1\omega_2
So, if we take the rotation to be almost all in the 1-axis (e.g. omega2 and omega 3 are small), and work to first order
We find that omega1 is roughly constant because we neglect the second order in smallness for the omega2*omega3 term; however, for omega 2 term (for example):
\frac{d^2}{dt^2}\omega_2=\left[\frac{(I_3-I_1)(I_1-I_2)}{I_3 I_2}\omega_1^2\right]\omega_2 (hopefully I did the algebra right).
We see then that for the omega2 term to remain small (and oscillate), then I1 must either be the largest or the smallest moment of inertia (thereby making the coefficient on the right hand side negative). If I1 is an intermediate moment of inertia, the coefficient on the right hand side is positive, and that means the solution for omega2 is exponential growth and not oscillatory.
One should note, though, that if you can PERFECTLY make the rotation ONLY on the 1-axis (with omega2 and omega3 being identically 0), then even if the 1-axis is the unstable axis, you won't get any wobbling. But this only works if you can make the rotation ONLY around the 1-axis.
