A calculus-free argument suggested by fedja:
The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is just of size $n^{-1/2}\ll \frac{1}{\sqrt{n}}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument (as suggested by fedja) using the fact that an $n$-element set has $2^n$ subsets. :
At least $n/2$ of the coordinates of any a point in the unit ball are at most $\sqrt{2/n}$ or less \sqrt{\frac{2}{n}}$ in absolute value, and the rest are at most $1$ in absolute value. Thus, the unit ball can be covered by at most $2^n$ bricks (meaning right-angled parallelepipeds) of volume $(8/n)^{n/4}$ $\left(2\sqrt{\frac{2}{n}}\right)^{n/2},$$ each . Each brick corresponds corresponding to one choice of the a subset of for the small coordinates. So Hence, the volume of the unit ball is at most $2^n $2^n \cdot (8/n)^{n/4} \left(2\sqrt{\frac{2}{n}}\right)^{n/2} = (128/n)^{n/4}$, which clearly goes to $0$.
Indeed\left(\frac{128}{n}\right)^{n/4}\rightarrow0.$$ In fact, the argument shows that the volume of the unit ball decreases faster than any exponential, which is to say that so the volume of the $n$-sphere ball of any fixed radius $r$ also goes to $0$.
