What you need is that E is not too singular, is bounded below, and grows sufficiently fast that the Boltzmann distribution is normalizable. The normalizability condition is the only thing, really.
The Boltzmann distribution is the stationary distribution for this Markov process, so you are guaranteed to converge to it if the Markov process is ergodic. This is a basic theorem of Markov processes, and it holds for continuous Markov processes as well. I'll give a unified proof below.
Discrete space, discrete time
There is a simple coupling argument that proves the convergence to the stationary distribution, in what I think is the best way, and it isn't presented often in books. It goes like this:
suppose X(t) is a Markov chain with an arbitrary starting point X_0 (or some starting distribution), and $Y(t)$ is the same process, but with the initial value $Y(0)$ picked from the stationary distribution. Define the coupled process as follows: if X(t) and Y(t) reach the same point, they move together from this point on, according to the Markov chain. If you ignore Y and look at X, this process is just the Langevin process on X. If you ignore X, it's the Langevin process on Y, so this is a coupling.
But in a discrete bounded state space, where the Markov chain has a finite probability of reaching any point from any other in a finite time (you can prove this by finding one path to do this), you can easily prove that there is always some chance for collision, so that the probability that the two walks couple goes to 1 at long times, and X is therefore in the stationary distribution.
Continuous space, discrete time
To prove the convergence to the Boltzmann distribution in continuous space but discrete time, you can do the following. This is the natural extension of the coupling on discrete state-spaces for continuous space:
Given two distributions $\rho_1$ and $\rho_2$, define their coupling distance $\Delta$ by
$$ 1-\Delta = \int \mathrm{min}(\rho_1(x),\rho_2(x)) $$
If $\Delta=0$, the two distributions are the same. If $\Delta=1$, the intersection of their support is measure zero.
The distributions $\rho_1$ and $\rho_2$ can be thought of as split into
$$ \rho_1 = \Delta \rho + (1-\Delta) (\rho_1 - \rho)$$
$$ \rho_2 = \Delta \rho + (1-\Delta) (\rho_2 - \rho)$$
Where
$$ \rho(x) = \mathrm{min}(\rho_1(x),\rho_2(x))$$
In other words, pick from the common distribution $\rho$ with probability $\Delta$, and from the residual distribution with probability $1-\Delta$.
Now if you have a discrete time continuous process on a finite dimensional state-space, usually you can prove the following for two nearby starting points: the coupling distance between the evolution of x and the evolution of y is nonzero when x and y are close.
When this is so, then you can start off X and Y as before, with Y from the stationary distribution, and each time step, you make x(t) equal to y(t) with probability $\Delta$, where $\Delta$ is the coupling distance between the evolution of x(t) and the evolution of y(t).
Continuous processes continuous time
In this case, you need a short-time estimate on the Fokker-Planck equation. You convert the process to discrete time, and you couple the discrete time versions as above. You need to show that if you start at x and start at y, and wait a little bit, you get a continuous distribution around x and around y. This means that the coupling distance is nonzero for close enough points, and then if your space is finite dimensional and path connected, you can prove that the stationary distribution is unique by the method above.
There is no assumption of high dimensionality--- the Langevin equation always produces a pick from the Boltzmann distribution, when it is normalizable. I am sure you can find these results proved in many places, with varying degree of rigor, but I think the method sketched above is optimal.
