Why does the expected norm of an $n$-dimensional vector grow like $\sqrt n$? Let $X_1, \ldots, X_n$ be i.i.d. standard normal r.v. According to this answer, for the expected Euclidean norm of the random vector $X:=(X_1, \ldots, X_n)^T \in\mathbb R^n$ holds the following:

$$\frac{n}{\sqrt{n+1}}\le E(\|X\|)\le \sqrt{n}.$$

So $E(\|X\|)$ asymptotically equals to $\sqrt n$.
Is there some illuminating geometric explanation for that?

It's the Pythagorean theorem, more or less.  The quantity $\|X\|^2$ is the sum of the squares of the components, each of which is roughly 1 in magnitude.  So $\|X\|^2$ is roughly $n$, and hence $\|X\|$ is roughly $\sqrt n$.  One can use the Weak  Law of Large numbers to conclude that the ratio $\|X\|/\sqrt n$ tends to 1 in probability, and the Central Limit theorem to get more precise estimates how much $\|X\|$ deviates from $1$ .
And so on.
