How to prove that, for any integer $d \geq 1$ and any $\alpha > 1/2$,
$$\int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} dr$$ goes to $0$ with $n$?
This can be interpreted (up to a multiplicative constant) as the integral of a multivariate gaussian distribution over points having distance $> \sqrt{n}$ from the origin.
Let $u = r^2d/n$ so that
$$ \int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} \, \mathrm{d}r = \tfrac{1}{2} (n/d)^{d/2} \int_{n^{2\alpha-1}d}^{\infty} u^{(d-2)/2} e^{-u} \, \mathrm{d}u. $$
From the L'hospital's rule, we know that
$$ \frac{\int_{x}^{\infty} u^{(d-2)/2} e^{-u} \, \mathrm{d}u}{x^{(d-2)/2} e^{-x}} \sim 1 \quad \text{as } x \to \infty. $$
Therefore we have
\begin{align*} \int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} \, \mathrm{d}r &\sim \tfrac{1}{2} (n/d)^{d/2} \cdot (n^{2\alpha-1}d)^{(d-2)/2} e^{-n^{2\alpha-1}d} \\ &= c n^{\beta} e^{-n^{2\alpha-1}d} \end{align*}
for $c = 1/(2d)$ and $\beta = \alpha(d-2)+1$. Taking limit as $n \to \infty$ gives the desired result.