I had been thinking to ask this same question, but I'll try my tentative answer.
First, by multiplying the equation by $-zF$ the voltages can be converted to energies for the reaction (rather than the electron):
$$\Delta G = \Delta G^\circ + RT \ln Q$$
The gas constant R is the Boltzmann constant multiplied by Avogardo's number divided by mol. "Six of one and half a dozen of the other", in other words. The gas constant and Boltzmann constant are essentially the same thing in different units.
So when we're looking at this equation, we see an energy associated with each concentration on the top and the bottom of the reaction quotient, each representing one particle that takes part in the reaction as a reactant or as a product.
To keep things "simple" in the following explanations, let's consider the "reaction" in which there is a hole in a membrane that allows a certain kind of ion to pass through - these ion channels are used widely in biology, such as to make a high-energy compound (ATP) in your cells. Neglecting the interaction between these ions, all the potential energy used to make that ATP comes out of the difference in concentration and its effect on this term of the Nernst equation. The cell potential (EMF) for this "reaction" is zero if the ion is at the same concentration on both sides of the membrane.
Notice that the energy comes from the quotient: it doesn't matter if the ratio is 10 M to 1 M or 0.1 M to 0.01 M. This is because we write this equation for just one channel, and so we're considering only the odds the ion goes through it one way or the other.
We also don't care about the rate of the reaction, because this is pure thermodynamics. We could write a Nernst equation for turning lead to gold and leave the philosopher's stone as an exercise for the student. There would still be some equilibrium potential you could calculate - though I'm not sure what the standard state is for neutrinos. ;)
Now the energy of an ion passing through an ion channel is going to average (rms) $\frac{1}{2} mv^2$ in one direction, and equal $\frac{1}{2} k_B T$. Notice this can be much less than the $\ln$ expression times $k_B T$ (i.e. $RT$ for one particle) that the Nernst equation gives.
My interpretation of this, which I want to check here, is that this happens because of the randomness of individual particle energy in the Boltzmann distribution. Certainly the energy predicted by the Nernst equation is dictated by that distribution: $\exp(-E/k_B T)$.
If an ion channel faces out to a tank of distilled water, the ion passes out and cannot return. If you couple that to a very high energy reaction, what you are doing is waiting until the thermal energy in the region happens to all coalesce and come together at a single point to make that reaction go forward, which then allows one ion to leave. As a result, the reaction cannot reverse unless that ion finds its way home. What's a bit remarkable about that interpretation is that we seem to see entropy reversing - pure heat energy being sucked into a chemical or a high-energy electrons and bottled up. This happens because we pay at least as much entropy in letting the ion loose into a dilute reservoir.
As we approach infinite energy from the gradient, kinetic effects will dominate. The rarity of sufficiently energetic particles drops off exponentially, at the $\exp(-E/k_BT)$ rate. Some measurement problems might be the leakage current of a voltmeter or the tendency of ions to collide with the solvent, diffuse slowly, and remain at a much higher concentration near the channel. Neither of these should invalidate the Nernst equation at the theoretical level; they just make it increasingly hard to observe.
