Consider this: there is an inverse relationship between the volume and the temperature of a gas. The air passing through your first stage valve expands dramatically. All other things equal (and they aren't, at least not in your regulator, where humidity, materials and frictive heating confuse the issue) the temperature drop can exceed 10F for every 1/4 ata. If the supply cylinder is at 3000 psi and the IP is set to 140psi, you're looking at a whole lot of adiabatic cooling.
This is all entirely wrong, I am sorry. Expansion cooling occurs only if entropy is kept constant, that means a fully reversible expansion process. A gas turbine goes quite close to it.
But in a lamination valve (as the first stage of a regulator) no work is obtained by the gas expansion, hence the entropy grows significantly. What remains almost perfectly constant is the enthalpy.
Assuming air is a perfect gas (which is almost perfectly, but not entirely), the temperature is directly proportional to entalpy: H = cp*T . Hence, as enthalpy is constant, also the temperature will not change.
The only way the expanding gas has to transform part of its enthalpy into another form energy is by converting to kinetic energy, that, is creating a very fast air jet.
That's the reason for, while breathing normally (hence with low speed of gas), the reg does not freeze. Instead, when you press the purge button creating a very fast flow, then some energy is transformed from enthalpy to kinetic, hence the expansion becomes more adiabatic, and the temperature drops.