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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 $n^{-1/2}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument using the fact that an $n$-element set has $2^n$ subsets. At least $n/2$ of the coordinates of any point in the unit ball are $\sqrt{2/n}$ or less and the rest are at most $1$ in absolute value. Thus, the ball can be covered by at most $2^n$ bricks (meaning right-angled parallelepipeds) of volume $(8/n)^{n/4}$ each. Each brick corresponds to one choice of the subset of small coordinates. So the volume of the ball is at most $2^n \cdot (8/n)^{n/4} = (128/n)^{n/4}$, which clearly goes to $0$.

Indeed, the volume decreases faster than any exponential, which is to say that the volume of the $n$-sphere of any fixed radius $r$ also goes to $0$.

The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is of size $\frac{1}{\sqrt{n}}\ll 1$. This can be turned into a 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 a point in the unit ball are at most $\sqrt{\frac{2}{n}}$ 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 (right-angled parallelepipeds) of volume $$\left(2\sqrt{\frac{2}{n}}\right)^{n/2},$$ each corresponding to a subset for the small coordinates. Hence, the volume of the unit ball is at most $$2^n \cdot \left(2\sqrt{\frac{2}{n}}\right)^{n/2} = \left(\frac{128}{n}\right)^{n/4}\rightarrow0.$$ In fact, the argument shows that the volume of the unit ball decreases faster than any exponential, so the volume of the ball of any fixed radius also goes to $0$.

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 $n^{-1/2}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument using the fact that an $n$-element set has $2^n$ subsets. At least $n/2$ of the coordinates of any point in the unit ball are $\sqrt{2/n}$ or less and the rest are at most $1$ in absolute value. Thus, the ball can be covered by at most $2^n$ bricks (meaning right-angled parallelepipeds) of volume $(8/n)^{n/4}$ each. Each brick corresponds to one choice of the subset of small coordinates. So the volume of the ball is at most $2^n \cdot (8/n)^{n/4} = (128/n)^{n/4}$, which clearly goes to $0$.

Indeed, the volume decreases faster than any exponential, which is to say that the volume of the $n$-sphere of any fixed radius $r$ also goes to $0$.

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 $n^{-1/2}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument using the fact that an $n$-element set has $2^n$ subsets. At least $n/2$ of the coordinates of any point in the unit ball are $\sqrt{2/n}$ or less and the rest are at most $1$ in absolute value. Thus, the ball can be covered by at most $2^n$ bricks (meaning right-angled parallelepipeds) of volume $(8/n)^{n/4}$ each. Each brick corresponds to one choice of the subset of small coordinates. So the volume of the ball is at most $2^n \cdot (8/n)^{n/4} = (128/n)^{n/4}$, which clearly goes to $0$.

Indeed, the volume decreases faster than any exponential, which is to say that the volume of the $n$-sphere of any fixed radius $r$ also goes to $0$.

The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is just of size $n^{-1/2}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument if we are allowed to use the simple combinatorial fact that and $n$-element set has $2^n$ subsets. Indeed, at least $n/2$ of the coordinates of any point in the unit ball are $\sqrt{2/n}$ or less and the rest are at most $1$ in absolute value. Thus, the ball can be covered by at most $2^n$ parallelepipeds of volume $(8/n)^{n/4}$ each (each parallelepiped corresponds to one choice of the subset of small coordinates). So the volume of the ball is at most $2^n \cdot (8/n)^{n/4} = (128/n)^{n/4}$, which clearly tends to $0$.

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 $n^{-1/2}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument using the fact that an $n$-element set has $2^n$ subsets. At least $n/2$ of the coordinates of any point in the unit ball are $\sqrt{2/n}$ or less and the rest are at most $1$ in absolute value. Thus, the ball can be covered by at most $2^n$ bricks (meaning right-angled parallelepipeds) of volume $(8/n)^{n/4}$ each. Each brick corresponds to one choice of the subset of small coordinates. So the volume of the ball is at most $2^n \cdot (8/n)^{n/4} = (128/n)^{n/4}$, which clearly goes to $0$.

Indeed, the volume decreases faster than any exponential, which is to say that the volume of the $n$-sphere of any fixed radius $r$ also goes to $0$.