Using the formula for the pdf of the Irwin-Hall 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 will have coordinates that 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 $\epsilon \rightarrow 0$
The first values are, $1,2\sqrt{2}, 3\sqrt{3}, \frac{32}{3}, \frac{115\sqrt{5}}{12},\frac{88\sqrt{6}}{5},\ldots$
This paper proposes an algorithm for a slight generalisation