By a strange coincidence I found myself thinking about almost this very question last week on a walk home early one morning (yes, that's correct). I wanted however only the weaker result that the ratio of the volume of the unit ball in the l^2 $l^2$ norm of R^n $\mathbb{R}^n$ to the that of the unit ball in the l^{infinity} $l^{\infty}$ norm of R^n $\mathbb{R}^n$ (i.e., 2^n), $2^n$), goes to zero has as $n -> oo. \to \infty$. This is what I came up with:
Start with the volume of the 4-ball in R^4. $\mathbb{R}^4$. Notice that the region is entirely contained in
that of the polydisk = {(x_1,x_2,x_3,x_4) $\{(x_1,x_2,x_3,x_4) | x_1^2 + x_2^2 <= \le 1, x_3^2 + x_4^2 <= 1} \le 1\}$ and therefore the proportion of the volume of the 4-sphere to that of the unit ball in the l^{\infty} $l^{\infty}$ norm is at most (pi/4)^2. $(\pi/4)^2$. Repeating this argument shows that the corresponding proportion in 2n $2n$ or 2n+1 $2n+1$ variables is at most (pi/4)^n. $(\pi/4)^n$. This goes to 0 as $n -> infinity.\to \infty$.
ADDENDUM: Having thought about it a little more, the above "polydisk" argument can be easily modified to answer the orginal question provided you exhibit one value of n for which the volume of the n-ball is less than one. The good news is that finding such an n is a straightforward calculus exercise involving repeated integration by parts. The bad news is that, if I've done the exercise correctly, the smallest such n is n = 13.
