Why does vibrational cooling occur much faster than sponteneous emission in the condensed phase?
As you point out, there are many places for this relatively small amount of vibrational energy to go. There are both radiative and non-radiative transitions which can take place to vibrationally quench the system in the electronic excited state. The explanation for both parts of your question (the vibrational and electronic parts) can be understood by Fermi's Golden Rule. I won't go into a lot of detail of deriving this or anything else because you can find anything you wanna know about it by searching online.

Fermi's golden rule says that the transition rate, $\Gamma_{i\to f}$, from some initial state, $|i\rangle$, to a set of final states, $|f\rangle$, is given by,
$$
\Gamma_{i\to f}=\frac{2\pi}{\hbar}|\langle f|H'|i\rangle|^2\rho
$$
where $\rho$ is the density of final states of the system and $H'$ is a perturbing Hamiltonian of some kind. For instance, the electric field of a photon. This is what it would usually be for absorption. In the case of decay, it may also be the field of a photon, but I am not as familiar with using Fermi's golden rule in this case.

Notice that this rule has dimensions of inverse time. It can be shown that the inverse of this rate is equal to the mean lifetime of the excited state (see this page).

Basically all of the physics you are asking about is contained within the fact that this transition rate is proportional to the density of possible states to which the system will decay. To put it simply, there are many many more intramolecular and intermolecular transitions which can take place to redistribute this energy. The vibrational density of states for this problem could probably be approximated quite well as being continuous because there will be many vibrational levels which are slightly perturbed due to distortions from equilibrium geometries. Think of the width of vibrational bands in IR.

On the other hand, a very large amount of energy must be dissipated for electronic relaxation to take place, and the density of these high-energy states is very low. Especially if one excites to the lowest electronic state. If, however, excitation occurs to a higher electronic energy level, decay should take place faster because the density of electronic states becomes much larger.

These processes have almost nothing to do with the speed of nuclear motion compared to electronic motion, as the condition which must be satisfied is really conservation of energy, and this is not always easy to do when the energy levels are quantized. From this perspective, the result is not surprising because vibrational energy levels are certainly closer to being continuous than electronic energy levels. Also, the spacing of vibrational state is generally smaller on an excited state than the ground state, so this adds to the quickness of the decay to the ground vibrational state.



As an aside, this paper is of some interest as it shows that vibrational relaxation times in the gas phase (for this one molecule) are slower than vibrational relaxation in solution. This can also be understood by the density of states argument discussed above.