For a conceptual proof, (1.) do not focus on the exact values of the integrals involved and (2.) turn to polar coordinates.
All norms on $\mathbb R^2$ are equivalent and, if $(x,y)=(r\cos\theta,r\sin\theta)$, then $r$ is the $\ell^2$ norm of $(x,y)$ while $|x|+|y|$ is its $\ell^1$ norm hence $ar\leqslant|x|+|y|\leqslant br$ for some absolute constants $a$ and $b$ (and it happens that $a=1$ and $b=\sqrt2$ but these values are anecdotal here). For every positive $c$, the $L^p(\mathbb R^2)$ norm of $(x,y)\mapsto1/(1+cr)$ is $ \iint\frac{r\mathrm dr\mathrm d\theta}{(1+cr)^p}=2\pi\int_0^{+\infty}\frac{r\mathrm dr}{(1+cr)^p}. $ The function in the last integral is locally integrable and equivalent to a multiple of $r^{1-p}$ at infinity, hence it is integrable if and only if $1-p\lt-1$, that is, $p\gt2$. Since this does not depend on $c$, the same holds for the original function of $(x,y)$.
Likewise, in $\mathbb R^d$, the function $(x_1,\ldots,x_d)\mapsto1/(1+|x_1|+\cdots+|x_d|)$ is in $L^p(\mathbb R^d)$ if and only if $p\gt d$.