Take a unit cube (of side 1) in N dimensions. Construct the crosssection at the midpoint of the longest diagonal. What is the area of this N1 dimensional region? I can compute this, but it would be nice to have a reference and a formula to crosscheck.

$\begingroup$ do you mean the crosssection which is perpendicular to the longest diagonal? $\endgroup$ – Yemon Choi Apr 26 '10 at 21:45

$\begingroup$ @Ila: You might expand the problem by posing your solutions (at least the related integral). There could be references for (more general!) formulas like yours without the geometric background. $\endgroup$ – Wadim Zudilin Apr 26 '10 at 22:44
For $n$ even, I get $$\frac{\sqrt{n}}{(n1)!} \left\langle \begin{matrix} n1 \\ n/21 \end{matrix} \right\rangle.$$
Here $\left\langle \begin{matrix} a \\ b \end{matrix} \right\rangle$ is the Eulerian number: the number of permutations of $a$ elements with $b$ descents.
Proof sketch: We are dealing with the convex hull of the $\binom{n}{n/2}$ zeroone vectors which have $n/2$ zeroes and $n/2$ ones. This is better known as the $(n/2,n)$hypersimplex. The hypersimplex is well known to have a triangulation with $\left\langle \begin{matrix} n1 \\ n/21 \end{matrix} \right\rangle$ tetrahedra. There are many references for this; my favorite, which discusses a number of prior references, is the early parts of Lam and Postnikov.
All the tetrahedra in this triangulation have the same volume, which is the volume of the convex hull of $e_1$, $e_2$, \ldots, $e_{n1}$. If I didn't screw up, that volume is $\frac{\sqrt{n}}{(n1)!}$.
This is a very old problem and there is a classical analytic approach to it. You can express the volume of sections of a convex body in terms of the Fourier transform of powers of the Minkowski functional.
Let $Q_n=[1/2,1/2]^n$ be the unit cube in $\mathbb R^n$ and let $[\xi^\perp]$ denote the hyperplane orthogonal to the vector $\xi\in S^{n1}$. Then $$Vol_{n1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{\infty}^{\infty} \prod\limits_{k=1}^n\frac{\sin(r \xi_k)}{r\xi_k}dr.$$
In our case $\xi=\xi_*=n^{1/2}(1,1,\dots,1)$ so the integral becomes $$Vol_{n1}(Q_n\cap [\xi_*^\perp])=\frac{1}{\pi}\int_{\infty}^{\infty} \left(\frac{\sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$ In fact, the latter formula was already known to Laplace! The integral tends to $\sqrt {6/\pi}$, as $n\to\infty$ (e.g. by Laplace's method). This can be also justified via the probabilistic interpretation suggested by Michael Lugo.
The Fourier analytic approach to sections of convex bodies is nicely presented in The Interface between Convex Geometry and Harmonic Analysis by A. Koldobsky and V. Yaskin. The derivation of the formula for volumes is available here.
EDIT (15.01.2011). In fact both integrals can be calculated explicitly. The sinc integrals were studied by Borwein & Borwein (see also MultiVariable sinc Integrals and the Volumes of Polyhedra).
For any $n\in \mathbb N$, $n>1$, we have $$\frac{1}{\pi}\int_{\infty}^{\infty} \left(\frac{\sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr =\frac{\sqrt n}{2^{n1}(n1)!}\sum_{k=0}^{n/2}(1)^k{n \choose k}(n2k)^{n1}$$

1$\begingroup$ Note that
$\int_{\infty}^{\infty} \ (sin(t)/t)^n dt$
can be computed exactly by residues for any specific positive integer $n$. $\endgroup$ – David E Speyer May 1 '10 at 14:05 
$\begingroup$ Thanks, David. I should have mentioned this in the answer. $\endgroup$ – Andrey Rekalo May 1 '10 at 22:28
Let the cube be $[0,1]^n$. I'm assuming you mean the crosssection perpendicular to the longest diagonal, from the allzeros corner to the allones corner. This is the crosssection corresponding to the plane $x_1 + \cdots + x_n = n/2$.
Alternatively, then, you want the probability density function of the sum $S_n$, of $n$ independent uniform(0,1) random variables, evaluated at $n/2$.
By the central limit theorem, the $S_n$ have asymptotic normal distribution with mean $n/2$ and variance $n/12$. So the answer should behave asymptotically like $1/\sqrt{2\pi n/12}$, i. e. like $\sqrt{\pi/6n}$.
It appears that this distribution is called the IrwinHall distribution.

$\begingroup$ Michael, I think that the answer is rather $6/\sqrt \pi$ for large n. $\endgroup$ – Andrey Rekalo May 1 '10 at 3:50

$\begingroup$ You're right  I didn't do the algebra correctly. $\endgroup$ – Michael Lugo May 1 '10 at 4:00

$\begingroup$ I'm sorry, the final answer is $\sqrt{6/\pi}$. $\endgroup$ – Andrey Rekalo May 1 '10 at 22:33