Voltage needed for electrolysis of water
I'm assuming you mean this Wikipedia article on the hydrolysis of water in which the method of determining that potential difference is described farther down the page.

I'll address your other questions - what might cause this value to change.


  Volume


No, except see below on concentration. The reaction potential for this electrolysis is independent of the amount of water since it is actually a thermodynamic function. Electrolysis may occur at different rates and take varying time depending on the amount of water, but the potential needed remains the same.


  Conductivity


Conductivity is a measure of the ease of transmitting electrical current (electrons) through a substance. So, increased conductivity will increase the rate of electrolysis, but since it does not change the original or final substances in the chemical equation, does not change any thermodynamic values like the reaction potential.

However, there is a danger here. We typically increase conductivity by increasing electrolyte content. Some of these electrolytes could be electrolyzed at a reduced potential compared to water. For example, if you use sodium iodide as your electrolyte, the iodide ion is oxidized to the triiodide ion at -0.53 V:

$$\ce{3I- -&gt; I3- + 2e-}\ \ E^\circ = -0.53\ \pu{V}$$


  Concentration


The reaction potential -1.23 V is the standard reaction potential, which is determined when all applicable species are at 1 molar concentration and 1 bar pressure. The two half reactions in this case are:

$$\ce{2H2O -&gt; O2 + 4H+ + 4e-}\\
\ce{4H+ + 4e- -&gt; 2H2}$$

Thus, this potential is only valid at $[\ce{H+}] = 1\ \pu{M}$ or $\pu{pH}=1$. As the reaction proceeds, water is consumed and the volume decreases. The amount of protons remains constant, so the pH increases. As the pH increases, so too does the potential required and the electrolysis eventually reaches equilibrium and progresses no further unless you are providing an overpotential.

Note that -1.23 V is equally valid at $[\ce{OH-}] = 1\ \pu{M}$ or $\pu{pH}=14$ thanks to Hess's Law.