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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 $\xi\in S^{n-1}$. Then $$Vol_{n-1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \prod\limits_{k=1}^n\frac{sin(r \xi_k)}{r\xi_k}dr.$$

In our case $\xi=\xi_*=n^{-1/2}(1,1,\dots,1)$ so the integral becomes $$Vol_{n-1}(Q_n\cap [\xi_*^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$ 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. As David noted, the second integral can be expressed in a fairly explicit form. For any $n\in \mathbb N$, $n>1$, we have $$\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr =$$ $$\frac{\sqrt n}{2^{n-1}(n-1)!}\left(n^{n-1}-n(n-2)^{n-1}+\frac{n(n-1)}{2!}(n-4)^{n-1} - \frac{n(n-1)(n-2)}{3!}(n-6)^{n-1}+...\right)$$

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 $\xi\in S^{n-1}$. Then $$Vol_{n-1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \prod\limits_{k=1}^n\frac{\sin(r \xi_k)}{r\xi_k}dr.$$

In our case $\xi=\xi_*=n^{-1/2}(1,1,\dots,1)$ so the integral becomes $$Vol_{n-1}(Q_n\cap [\xi_*^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{\sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$ 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 Multi-Variable sinc Integrals and the Volumes of Polyhedra). 

For any $n\in \mathbb N$, $n>1$, we have $$\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{\sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr =\frac{\sqrt n}{2^{n-1}(n-1)!}\sum_{k=0}^{n/2}(-1)^k{n \choose k}(n-2k)^{n-1}$$

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 $\xi\in S^{n-1}$. Then $$Vol_{n-1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \prod\limits_{k=1}^n\frac{sin(r \xi_k)}{r\xi_k}dr.$$

In your case $\xi=\xi_n=n^{-1/2}(1,1,\dots,1)$ so the integral becomes $$Vol_{n-1}(Q_n\cap [\xi_n^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$ Actually, the latter formula was already known to Laplace! The integral tends to $6/\sqrt \pi$, as $n\to\infty$ (e.g. by Laplace's method). This can be also justified via the probabilistic interpretation suggested by Michael Lugo.

This is the so called Fourier analytic approach to sections of convex bodies. I'd recommend Fourier Analysis In Convex Geometry by Alexander Koldobsky as a reference.

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 $\xi\in S^{n-1}$. Then $$Vol_{n-1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \prod\limits_{k=1}^n\frac{sin(r \xi_k)}{r\xi_k}dr.$$

In our case $\xi=\xi_*=n^{-1/2}(1,1,\dots,1)$ so the integral becomes $$Vol_{n-1}(Q_n\cap [\xi_*^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$ 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. As David noted, the second integral can be expressed in a fairly explicit form. For any $n\in \mathbb N$, $n>1$, we have $$\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr =$$ $$\frac{\sqrt n}{2^{n-1}(n-1)!}\left(n^{n-1}-n(n-2)^{n-1}+\frac{n(n-1)}{2!}(n-4)^{n-1} - \frac{n(n-1)(n-2)}{3!}(n-6)^{n-1}+...\right)$$

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 $Q_n$ be the unit cube in $\mathbb R^n$ and let $[\xi^\perp]$ denote the hyperplane orthogonal to the vector $\xi\in S^{n-1}$. Then $$Vol_{n-1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \prod\limits_{k=1}^n\frac{sin(r \xi_k)}{r\xi_k}dr.$$

In your case $\xi=\xi_n=n^{-1/2}(1,1,\dots,1)$ so the integral becomes $$Vol_{n-1}(Q_n\cap [\xi_n^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$ Actually, the latter formula was already known to Laplace! The integral tends to $6/\sqrt \pi$, as $n\to\infty$ (e.g. by Laplace's method). This can be also justified via the probabilistic interpretation suggested by Michael Lugo.

This is the so called Fourier analytic approach to sections of convex bodies. I'd recommend Fourier Analysis In Convex Geometry by Alexander Koldobsky as a reference.

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 $\xi\in S^{n-1}$. Then $$Vol_{n-1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \prod\limits_{k=1}^n\frac{sin(r \xi_k)}{r\xi_k}dr.$$

In your case $\xi=\xi_n=n^{-1/2}(1,1,\dots,1)$ so the integral becomes $$Vol_{n-1}(Q_n\cap [\xi_n^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$ Actually, the latter formula was already known to Laplace! The integral tends to $6/\sqrt \pi$, as $n\to\infty$ (e.g. by Laplace's method). This can be also justified via the probabilistic interpretation suggested by Michael Lugo.

This is the so called Fourier analytic approach to sections of convex bodies. I'd recommend Fourier Analysis In Convex Geometry by Alexander Koldobsky as a reference.