Using the formula for the pdf of the Irwin-Hall distribution one gets $$\frac{\sqrt{n} 2^{n-1}}{(n-1)!}\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}(-1)^{k}{n \choose k}\left(\frac{n}{2}-k\right)^{n-1}$$
This paper proposes an algorithm for a slight generalisation