Why is the zero of electric potential taken to be $r = \infty$, rather than $r = 0?$ Wouldn't it make more sense if it was taken to be zero at $r = 0$? This seems to imply that with a negative test charge at $r = 0$ from a positive point charge, $V = -\infty$, which I have trouble conceptualizing properly. Is that accurate? Is there a better way to look at it?
 A: You evidently understand that any constant can be added to a potential without affecting the physics — or equivalently, any place can be taken to have zero potential.  You also suggest, rightly, that there are really only two mathematically "natural" places to define the zero of the potential: either $r=\infty$ or $r=0$.  For example, there's no particular reason to choose $r=3$ whatever the units may be.  In fact, it is precisely because of this issue with units that the only two points that can be defined mathematically by a choice of $r$, independent of the choice of units, are $r=0$ and $r=\infty$.
Of course, it may well be true that the physics of some particular situation leads to a natural choice other than these two — for example, we frequently take the surface of some conductor as the zero of the potential.  But the question is with regards to generalities, so in general we're limited to just those two mathematical choices.
The problem with $r=0$, however, is that all of our usual approximations break down at that point.  In particular, the magnitude of the force in a standard inverse-square force law is infinite at that point (and its direction is indefinite).  To be more precise, in general you can say for any inverse-square force law that the potential is
\begin{equation*}
  V(r) = \frac{1}{r} + C,
\end{equation*}
where $C$ is some constant.  But $r=0$ gives you an infinite value.  And if you try to use $C$ to "subtract off the infinity", you'll get infinite values everywhere other than $r=0$.
Fundamentally, this is a problem in the relationship between the mathematical model and the physical model.  Mathematically, if you have a point particle coming towards an oppositely charged point particle (with no angular momentum), it will reach $r=0$ with infinite speed.  But physically, that's impossible.  That disagreement is pointing you to the fact that our simple mathematical model just isn't correct at very small distances; we need to change it to reflect physics if you want a better answer.  The best currently known way to deal with this is to use quantum field theory — which is more complicated mathematically, but also more correct physically.
A: This is accurate, and it ultimately comes down to the fact that we can get arbitrarily close to an electric point charge in classical E&M. That means that the field right up next to the point charge could be arbitrarily large. So you get these huge, singular potentials close to point charges, which is really more-or-less fine. For instance, that huge positive potential that's right next a positive point charge will keep other positive charges from getting close to it. 
In addition, often in electromagnetism vector algebra can reshuffle terms until we get something that depends on the potential "at the boundary surface" that encloses all the charge. If we can take this to infinity where $\phi = 0$, then we can drop that term, which is often convenient.
A: For a classical point charge, the field is divergent at $r=0$, and if you were to take the potential to be zero there, it would be infinite everywhere else.  Meanwhile, you can approximate $r=\infty$ as the region with no interaction, so it's reasonably naturally to treat it in the way you would treat ground.
