I came across the following problem and got stuck.
Problem: Let $X_1,X_2,...$ be independent Unif$(-1,1)$ and $S_n=X_1^2+...+X_n^2$. Let $$A_n=\{x\in \mathbb{R^n}:\sqrt\frac{n}{3}-1<\sqrt{x_1^2+...+x_n^2}<\sqrt{\frac{n}{3}}+1\}$$Show that for all sufficiently large $n$ the Lebesgue measure of $A_n\cap(-1,1)^n $ is at least $99\%$ of the Lebesgue measure of the cube $(-1,1)^n$.
Attempt: I have managed to show that $$\sqrt{S_n}-\sqrt{\frac{n}{3}}\xrightarrow{d}N(0,\sigma^2=\frac{4}{45})$$ using Slutsky's theorem and some algebra and observed that $$A_n\cap(-1,1)^n=\{x\in (-1,1)^n:\sqrt\frac{n}{3}-1<\sqrt{x_1^2+...+x_n^2}<\sqrt{\frac{n}{3}}+1\}$$and notice that this is just (if we assume $x_i$'s are uniformed distributed) $$A_n\cap(-1,1)^n=\{|\sqrt{S_n}-\sqrt{\frac{n}{3}}|<1\}$$ since we have shown that $\sqrt{S_n}-\sqrt{\frac{n}{3}}\xrightarrow{d}N(0,\sigma^2=\frac{4}{45})$, we have $$P(A_n\cap(-1,1)^n)\rightarrow P(|M|<1)$$where $M$ is normally distributed with mean $0$ and variance $\frac{4}{45}$. This probability is almost one (0.9999). And I don't know how to go on from here.