The following result is from Tao's note: https://terrytao.wordpress.com/2015/10/23/275a-notes-3-the-weak-and-strong-law-of-large-numbers/comment-page-1/#comment-682897.
Proposition: Let ${\varepsilon > 0}$. Then, for sufficiently large ${n}$, a proportion of at least ${1-\varepsilon}$ of the cube ${[-1,1]^n}$ (by ${n}$-dimensional Lebesgue measure) is contained in the annulus ${\{ x \in {\bf R}^n: (1-\varepsilon) \sqrt{n/3} \leq |x| \leq (1+\varepsilon) \sqrt{n/3} \}}$.
Proof: Let ${X_1,X_2,\dots}$ be iid random variables drawn uniformly from ${[-1,1]}$. Then the random vector ${(X_1,\dots,X_n)}$ is uniformly distributed on the cube ${[-1,1]^n}$. The variables ${X_1^2,X_2^2,\dots}$ are also iid, and (by the change of variables formula) have mean
$\displaystyle \int_0^1 x^2\ dx = \frac{1}{3}$.
Hence, by the weak law of large numbers, the quantity ${\frac{X_1^2+\dots+X_n^2}{n}}$ converges in probability to ${\frac{1}{3}}$, so in particular the probability
$\displaystyle {\bf P}(|\sqrt{X_1^2+\dots+X_n^2} - \sqrt{n/3}| > \varepsilon \sqrt{n})$
goes to zero as ${n}$ goes to infinity. But this quantity is precisely the proportion of ${[-1,1]^n}$ that lies outside the annulus ${\{ x \in {\bf R}^n: (1-\varepsilon) \sqrt{n/3} \leq |x| \leq (1+\varepsilon) \sqrt{n/3} \}}$, and the claim follows. $\Box$
Question: Isn't the proportion which lies outside the given annulus be $\displaystyle {\bf P}(|\sqrt{X_1^2+\dots+X_n^2} - \sqrt{n/3}| > \varepsilon \sqrt{n/3})$ instead?