Using the formula for the pdf of the [Irwin-Hall][1] distribution one gets
$$\frac{\sqrt{n}}{(n-1)!}\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}(-1)^{k}{n \choose k}\left(n-2k\right)^{n-1}$$

It's fairly straightforward to see why, imagine you're drawing random point in your cube, how many sum to less than $\epsilon$ in absolute value? This gives you a $2\sqrt{n}\epsilon$ thick slice of hypercube around the hyperplane. Take the limit as $\epsilon \rightarrow 0$



[This paper][2] proposes an algorithm for a slight generalisation 


  [1]: http://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution
  [2]: http://www.dcc.ufla.br/infocomp/artigos/v8.4/art03.pdf