We know that water auto-ionizes (to a rather small extent).
My question is which solvent would have the fewest ions (or at least fewer ions than water).
I'm thinking that perhaps some kind of hydrocarbon, as these are both weaker acids and weaker bases than water and thus even less reactive. I've seen computers submerged in oils (as well as distilled water) due to their low conductivities.
2-Methylpropane has a $\mathrm{p}K_\mathrm{a}$ of 53 ($\mathrm{p}K_\mathrm{a}$ table). That sounds pretty non-ionic. Can't think of any common all-carbon (no substituents of any kind) liquid compounds.
Ron's answer is a good option, however autoionization is not necessarily equivalent to self-protonation. There may be some solvents where autoionization is even more suppressed than any self-protonation reaction. Perhaps the liquid with the lowest tendency to ionize may be something composed of molecules with extremely strong bonds, for example liquid nitrogen or liquid carbon monoxide. Of course, in Ron's acid table there already are several compounds which would be expected to ionize less than once in even reasonably macroscopic volumes of liquid.
Edit: As Greg points out in the comments, why not go all out and consider noble gasses. The only plausible ionization reaction for the pure liquid elements is:
$$\ce{Ng(l) <=> Ng+_{(Ng)} + e-_{(Ng)}}\ ,$$
where $\ce{Ng}$ indicates a noble gas and the $\ce{(Ng)}$ subscript indicates solvation by the noble gas atoms. The electron doesn't attach to anything, as $\ce{Ng-}$ would actually be unbound (no barrier to dissociation, or at least this is true for non-ultraheavy noble gasses such as ununoctium and beyond).
Given that noble gas atoms likely make for an extremely poor solvent, we can get a very rough measure of the equilibrium constant by assuming the solvation is nil. Thus, the reaction would be equivalent to the gas-phase ionization of the atoms. Assuming the entropy change is a small factor relative to the very large ionization enthalpies, then the reaction $\Delta G_\mathrm{ion}^\circ$ is simply equal to the ionization energy.
For helium, this equals a massive $\pu{2372.3 kJ mol-1}$. Applying $\Delta G^\circ = -RT\ln K_\mathrm{ion}$, then for liquid helium at its boiling point under atmospheric pressure ($\pu{4.2 K}$), the ionization constant is as low as $10^{-29500}$ (though most of the reason the exponent is tiny is because of the really low temperature; for comparative purposes the value at room temperature would be around $10^{-400}$). A more accurate calculation would drive this value up quite a bit, but it would still be many, many orders of magnitude more difficult to ionize than anything else.
