How long for a frictionless top to fall over? We've previously discussed why it is that spinning tops do not fall over, see: Why don't spinning tops fall over?
However, as the highest rated answer notes, the angular momentum of the spinning top is "quite high". On the other hand, I know that if the angular momentum of the top is zero it will, in fact, fall over. This suggests that increasing the angular momentum cannot stop the top from falling over, but instead can only increase the time (but this turns out not to be the case).
So suppose a top with mass concentrated in a spherical body M of radius S with center situated a distance R to a point a horizontal table at an angle $\theta$ while rotating at angular speed $\omega$. How long does it take for the top to fall over?
Some simplifying assumptions: (1) Assume the point of the top is fixed on the table. (2) Assume that the time at which the top falls over is when $\theta = 0$.

 A: Given a nonzero moment of inertia $I$, the transverse extension of the top from the rod has to be at least $\sqrt{I/M}$ because we can't have negative mass distributions. Because of this finite transverse extension, a spin-orbit cross coupling between the orbital angular momentum $L$ and the spin $S$ will have to exist. But this by itself isn't sufficient to transfer spin into orbital angular momentum.
The crucial question to ask is, is the distribution of mass of the rod rotationally symmetric about the rod axis? If yes, the kinetic and gravitational potential energy doesn't depend upon the orientation angle $\alpha$ of the top about the rod axis. However, if there are slight rotational asymmetries, we will have an $\alpha$ dependence leading to wobbles. These wobbles transfer spin into orbital angular momentum, and if sufficient angular momentum is transfered, the top will topple.
A: I think, once launched, the top will never fall over. Such are kinematic restrictions in this problem. For example, $\theta = const$ if it exceeds the minimum angle necessary for the top to be detached from the table. Of course, I imply that the initial conditions on the angular momentum are such that $\theta > \theta_{min}$.
A: Well if you are really interested in Gyroscopes what about the equations of motion to get us started? Here I have copied some from http://www.gyroscopes.org/math2.asp which are in Euler-like coordinates: 
q = nutation angle - nutation rate q'= n
f = precession angle - precession rate f'= p
y = spin angle - spin rate y'= s
and primes denote differentiation wrt time. We are assuming that s is constant - this is because there is no friction.
$$SM_x=I_xn'-I_yp^2\cos q\sin q+I_zp\sin q(p \cos q+s)$$
$$SM_y=I_y(pn\cos q)-I_zn(p\cos q+s)+I_xpn \cos q$$
$$SM_z=I_z(-pn\sin q)$$
The expression for $SM_z$ simplifies because of the symmetry in the moments of inertia which in our case with the sphere on a rod R are:
$$I_x=mR^2+2/5mr^2$$
$$I_y=mR^2+2/5mr^2 = I_x$$
$$I_z=2/5mr^2$$
Now this is an external torque gyroscope - as we are on a table on some gravitational environment. The external torque applies in the x direction
$$F_x=mgR\sin q$$
Note that all of these equations are written in terms of the nutation angle q. In these coordinates q=0 when it is upright and $q=90^0$ when it is fallen over (although this is an approximation since it falls over at $q=90-\arctan(r/R)$).
The intuition behind your question is that when there is no spin s=0, there is no precession p=0. In this case the "fall" corresponds to a nutation from a small angle $q_0$ down to $q=90^0$ - and this nutation is the only motion. As s>0 these equations will generate a non-zero precession p>0, also the nutation will now reach a maximum $q_{max}$. The requirement is that s is sufficiently large that $q_0-q_{max}<90^0$. If this condition is not met then the top will still hit the surface after some rotation (precession angle f in these coordinates).
If one wished to take this further analytically then we could approximate for small initial angle $q_0$. So $\sin q_0=q_0$ and $\cos q_0=1$. Then we consider the x axis terms for the external torque.
$F_x=M_x$ - which holds at all times
$mgRq_0=I_xn'-I_yp^2q_0+I_zpq_0(p+s)$
This quadratic in p will determine it at t=0 in terms of other parameters (but n' might not be a constant - from graphical solutions I expect it to be sinusoidal). Likewise one can determine the "falling over" equation when $q=90^0$. If p=p(s) does not satisfy the "falling over" equation then the gyroscope will continue without falling. If it does we can integrate to determine q and hence q(t).
EDIT: Additional point on Gyroscope Intuition
Comments suggest that the motion described by the equations is not directly in accord with intuition. This is likely to be because a real Gyroscope has some elements of Friction. This friction has the effect of a damping term in the above equations. The friction will be basically a function of s, but n=n(s) through the equations, and the only acceleration term here is n'. Thus the friction, for a realistic Gyroscope, will damp the Nutation oscillations that otherwise occur. This would make the nutation an approximate constant of motion.
Treating nutation as a constant, simplifies the equations by setting n'=n=0 and so q, cos q and sin q become constants. However from an intuition perspective it reduces the number of (Euler) degrees of freedom from 3 to 2. So in a real Gyroscope we might see only the two degrees of freedom: spin and precession. Trying to intuit from this the behaviour in the remaining nutation dimension could be misleading as it ordinarily does not display behaviour in that dimension.
A related physical system which does display nutation is the "spinning plate on a pole" scenario. As the spin of the plates reduces through friction they start to wobble. If a wobble is too great the plate will fall over, even although it is still spinning a little.
