Atmospheric pressure will have some effects on the efficiency, even at pressures more moderate than the extreme cases Evan mentioned.

Two effects come to mind immediately (and there may be more):

The first is a thermodynamic effect--the atmosphere contains oxygen, which is a product of water electrolysis. By increasing the atmospheric pressure, you also increase the amount of oxygen that remains dissolved in the electrolyte, and this makes the production of oxygen slightly less favorable (a higher voltage is required). The voltage required to drive the oxygen production reaction (as with all electrochemical half-reactions) is related to the concentrations of product and reactant (see:

http://en.wikipedia.org/wiki/Nernst_equation)

For the reaction 2 H

_{2}O → O

_{2} + 4 H

^{+} + 4 e

^{–}E = E

^{0} – (R*T*ln([H

_{2}O]

^{2}*[O

_{2}]

^{–1}*[H

^{+}]

^{–4}))/F

where R is the universal gas constant, T is absolute temperature, F is Faraday's number, and [something] denotes the concentration of something.

The second effect is more of an engineering consideration--the higher the pressure is, the smaller the volume of the bubbles produced (inverse proportionality). This means that for a given rate of hydrogen and oxygen production, less of each electrode will be covered by bubbles (which are effective insulators, and block the reaction). Or, turned around, at higher pressures you can form more product per unit surface area than at low surface area.

The overall change in efficiency, given these competing effects, would be hard to quantify, as it is a function of the temperature, pressure, pH, electrode area, applied voltage, conductivity of the solution and probably a few more parameters I can't think of at the moment.

At reasonable pressures, both effects are minimal. If you want to increase the efficiency of your electrolysis substantially, the easiest variables to play with are composition of the electrolyte, electrode material, and cell geometry.