I think Adam's answer is excellent, and I'd rather make this a comment but can't as I just signed up in order to answer.
While I agree with most of what Adam said, there are cases where large-N works well for $N=1$. The case I'm familiar with is the large-N expansion for the "Spin Ices" Dy$_2$Ti$_2$O$_7$ and Ho$_2$Ti$_2$O$_7$, which have Ising spins on a pyrochlore lattice. The best reference for this is Sergei Isakov's PhD thesis (University of Stockholm 2004), specifically Section 4.3 and Subsection 4.8.2.
It's noted in that thesis that we shouldn't expect large-N to work for pyrochlores below $N=3$; the expansion is singular at $N=2$ owing to order-by-disorder. Extensive checking against Monte Carlo simulations and experiment verify that the expansion is good for $N=1$ (references given in the thesis), but we don't know why this is.
I found myself needing to prove the equivalence of large-N and MFT at high temperatures. I apologise if it's bad etiquette to reference one's own work, but I provide a mathematical proof in Section 2.4 of my Master's thesis which can be found here (Perimeter Institute 2011).
I believe the main working difference between large-N and MFT is that MFT always contains a non-zero critical temperature $T_c$, but that $T_c\sim 1/N$ with $N$ the number of degrees of freedom at each lattice site. This means that large-N, with $N\rightarrow\infty$, has $T_c=0$ to zeroth order in the $1/N$ expansion (at which one generally works).