Here's a geometric argument (still with a bit of calculus). The volume of the unit $n+1$-ball may be obtained from integrating the volumes of $n$-ball cross-sections from say south pole to north pole. We have$Vol(D^{n+1}) = Vol(D^{n}) \int_{-1}^1 \sqrt{1-z^2}^{n} dz$, since the volume of the $n$-ball of radius $r$ is the volume of the unit $n$-ball times $r^n$, and the radius of the $z$-cross-section is $\sqrt{1-z^2}$. Since for any $1 >\delta >0$, $(1-z^2)^{n/2}$ converges to $0$ uniformly on $[-1,-\delta] \cup [\delta, 1]$, it is not hard to see that these integrals converge to zero.
So the ratio $Vol(D^{n+1})/Vol(D^n)$ converges to zero, and therefore $Vol(D^n)\to 0$ as $n\to \infty$. As Gil Kalai says, this argument shows that the volume gets concentrated near the equator.