I assume only $\alpha \gt 1$ gives $\int_{\mathbb{R}^n} (1+|x|)^{-\alpha} \mathrm{d}x \lt \infty$ (simply because this is true for $n=1$). I also assume some clever transformation could be used for the proof.
I tried to proof this by induction to $n$. $n=1$ is quiet simple, for $n \mapsto n+1$ I thought about:
$$\Phi: ((0,\infty),(-\pi,\pi),\mathbb{R}^{n-2}) \rightarrow \mathbb{R}^n,\quad (r,\varphi,\overline{x}) \mapsto (r \cos \varphi, r \ sin \varphi, \overline{x}),$$
where $x = (x_1,x_2,\overline{x})$. Then $|\det \Phi'| = r$ and the transformation formula gives:
$$ \int_{R^n} (1+x)^{-\alpha} \mathrm{d}x = \int_{R^{n-2}} \int_0^\infty \int_{-\pi}^\pi \frac{r}{\left(1+\sqrt{r^2+(x_3^2+\dots+x_n^2)}\right)^\alpha} \mathrm{d}\varphi \mathrm{d}r \mathrm{d}(x_3,\dots,x_n) $$
The original idea was to somehow use the induction principle here, i. e. to somehow rip $\overline{x}$ out of $\sqrt{r^2+\overline{x}}$. But how?
Does this approach have a chance to be successful? If yes: How to follow up? If no: What else should I do?
PS: $|x|$ is the $2$-norm and all integrals are Lebesgue integrals.
Answer. $$ \int_{\mathbb R^n}\frac{dx}{(1+|x|)^a}<\infty $$ if and only if $a>n$.
Proof. The easiest way to show this is using polar coordinates: $dx=r^{n-1}dr\, d\vartheta$: $$ I(M)=\int_{|x|<M}\frac{dx}{(1+|x|)^a}=\int_0^M \left(\int_{|s|=1}\frac{1}{(1+r)^{a}} ds \right)r^{n-1}dr=\omega_{n-1}\int_0^M \frac{r^{n-1}\,dr}{(1+r)^a}. $$ where $\omega_{n-1}$ is the area of the unit sphere in $\mathbb R^n$. If $a>n$, then \begin{align} I(M)&\le \omega_{n-1}\int_0^M \frac{(1+r)^{n-1}\,dr}{(1+r)^a}=\omega_{n-1}\int_0^M (1+r)^{n-a-1}dr=\omega_{n-1}\left.\frac{(1+r)^{n-a}}{n-a}\right|_0^M \\&=-\frac{\omega_{n-1}(1+M)^{-a+n}}{n-a}+\frac{\omega_{n-1}}{a-n}<\frac{\omega_{n-1}}{a-n}. \end{align} Thus $$ \int_{\mathbb R^n}\frac{dx}{(1+|x|)^a}=\lim_{M\to\infty} I(M)\le \frac{\omega_{n-1}}{a-n}<\infty. $$ On the other hand, if $a\le n$, then \begin{align} I(M)&=\omega_{n-1}\int_0^M \frac{r^{n-1}\,dr}{(1+r)^a}>\omega_{n-1}\int_1^M \frac{r^{n-1}\,dr}{(1+r)^a}\ge \omega_{n-1}\int_1^M \frac{r^{n-1}\,dr}{(2r)^a} \\&= \frac{\omega_{n-1}}{2^a} \int_1^M r^{n-1-a}\,dr\ge \frac{\omega_{n-1}}{2^a} \int_1^M r^{-1}\,dr=\frac{\omega_{n-1}}{2^a} \log M\to\infty. \end{align}