In this answer, I will make my intuition of comparing the series to an integral rigorous.
From now on, $n$ always denotes any natural number and a vector $x\in\mathbb R^n$ or $x\in\mathbb Z^n$ will be written as $x=(x^1,\dots, x^n)$.
Theorem. For $p\in\mathbb R$, the sum $$\sum_{\mathbf k\in\mathbb Z^n\setminus\{0\}} \lvert\mathbf k\rvert^{-p}$$
converges (in the sense of the Lebesgue integral with respect to the counting measure) if and only if $p>n$, where $\lvert\mathbf k\rvert=\sqrt{\sum_{j=1}^n (\mathbf k^j)^2}$ denotes the usual Euclidean norm of $\mathbf k$.
Lemma 1. For any $n\in\mathbb N$ and $p\in\mathbb R$, $$\sum_{\mathbf k\in\mathbb Z^n\setminus\{0\}} \lvert\mathbf k\rvert^{-p}=2^n\sum_{\mathbf k\in\mathbb N^n}\lvert\mathbf k\rvert^{-p},$$
where both sides may be infinite.
Lemma 2. For $p\in\mathbb R$, the integral $$\int_{[1,\infty[^n} \lvert x\rvert^{-p}\,\mathrm dx,$$ in the Lebesgue sense, is finite if and only if $p>n$.
Lemma 3. For any $p\in\mathbb R$, $$\sum_{\mathbb k\in(\mathbb N\setminus\{1\})^n}\lvert\mathbf k\rvert^{-p}\le\int_{[1,\infty[^n} \lvert x\rvert^{-p}\,\mathrm dx.$$
Proof of the part of the Theorem where $p>n$ (by strong induction over $n$). By Lemma 1, the statement to prove is equivalent to $$\sum_{\mathbf k\in\mathbb N^{n}}\lvert\mathbf k\rvert^{-p}<\infty.$$
Base case ($n=1$). Follows directly from improper Riemann integration and the monotone convergence Theorem.
Induction step. Fix $n\in\mathbb N$ and suppose that $\sum_{\mathbf k\in\mathbb N^{m}}\lvert\mathbf k\rvert^{-p}<\infty$ for all $m\in\mathbb N\cap[1,n-1]$. Note $$\mathbb N^n =\bigcup_{X\subset\{1,\dots,n\}} \mathbb N_X,$$ where $$\mathbb N_X\overset{\text{Def.}}=\{\mathbf k\in\mathbb N^n:\mathbf k^x=1\text{ for all }x\in X\text{ and }\mathbf k^x\neq 1\text{ for all }x\not\in X\}.$$
By Lemma 2 and 3, if $p>n$, then $\sum_{\mathbf k\in\mathbb N_{\emptyset}} \lvert\mathbf k\rvert^{-p}<\infty$. By the induction hypothesis, for any $X\subset\{1,\dots,n\}$ which is not the empty set, $$\sum_{\mathbf k\in\mathbb N_{X}} \lvert\mathbf k\rvert^{-p}<\infty.$$
This proves that $$\sum_{\mathbf k\in\mathbb N^{n}}\lvert\mathbf k\rvert^{-p}<\infty.$$ $\square$
Sketch of the proof of the part of the Theorem where $p\le n$. Use the same argumentation as in Lemma 3 to show that $$\sum_{\mathbf k\in\mathbb N^n}\lvert\mathbf k\rvert^{-p}\ge\int_{[1,\infty[^n}\lvert x\rvert^{-p}\,\mathrm dx$$ and conclude using the same argumentation as in Lemma 2.
Proof of Lemmas
Proof of Lemma 1. We may write $$\mathbb Z^n\setminus\{0\}=\bigcup_{s=(s_1,\dots,s_n)\in\{-1,1\}^n} \underbrace{\prod_{j=1}^n (-1)^{s_j} \mathbb N}_{\overset{\text{Def.}}=A_s},$$
where $\prod$ denotes the generalized cartesian product. Now we note that, since $\lvert x\rvert$ is invariant under the changing of sign of any coordinate of $x$, $$\sum_{\mathbf k\in A_{s}}\lvert\mathbf k\rvert^{-p}=\sum_{\mathbf k\in A_{\tilde s}} \lvert \mathbf k\rvert^{-p}$$ for any $s,\tilde s\in \{-1,1\}^n$. (Formally, this can be justified using the Transformationssatz (cf. for instance [1; Theorem 1]) for the pushforward under the flipping of coordinates of $\mathbf k$.)
The right-hand side of the Lemma is therefore $$\sum_{s\in\{-1,1\}^n}\sum_{\mathbf k\in A_{s}}\lvert\mathbf k\rvert^{-p},$$ which therefore equals the left-hand side from linearity of the sum. $\square$
Proof of Lemma 2. We use once again the Transformationssatz. Consider the function \begin{equation*}\begin{split}\Phi:]0,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[&\to\mathbb R^n\\\begin{pmatrix}r\\\phi_1\\\phi_2\\\dots\\\phi_{n-2}\\\phi_{n-1}\end{pmatrix}&\mapsto\begin{pmatrix}r\cos(\phi_1)\\r\sin(\phi_1)\cos(\phi_2)\\r\sin(\phi_1)\sin(\phi_2)\cos(\phi_3)\\\dots\\r\sin(\phi_1)\dots\sin(\phi_{n-2})\cos(\phi_{n-1})\\r\sin(\phi_1)\dots\sin(\phi_{n-2})\sin(\phi_{n-1})\end{pmatrix}.\end{split}\end{equation*}
Exercise. Prove that there exists a Lebesgue-null set $N\subset\mathbb R^n$ such that
- $\Phi:]0,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[\to\mathbb R^n\setminus N$ is a $C^1$-diffeomorphism;
- for all $(r,\phi_1,\dots,\phi_{n-1})\subset]0,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[$, we have $$\left\lvert\operatorname{det}\operatorname{Jac}\Phi((]0,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[)^\top)\right\rvert=r^{n-1}\sin^{n-2}(\phi_1)\sin^{n-3}(\phi_2)\dots\sin(\phi_{n-2}).$$
- $\Phi(]1,\infty[\times]0,\pi[^{n-2}\times]0,2\pi[)\supset[1,\infty[^n$.
Hint. First, prove that $\Phi$ is injective. Second, note that, for a fixed $r$, the image of $\Phi$ is $r\mathbb S^{n-1}$, up to a Lebesgue-measurable set of Hausdorff dimension $n-2$. Then conclude that $\Phi$ is surjective up to a Lebesgue-measurable set of Hausdorff dimension $n-1$, i.e. of $n$-Lebesgue measure $0$. For the computation of the Jacobian, see https://en.wikipedia.org/wiki/N-sphere#Spherical_volume_and_area_elements.
Back to the proof of Lemma 2: From 1. and 3. above together with the Transformationssatz [1; Theorem 1] we get $$\int_{[1,\infty[^n} \lvert x\rvert^{-p}\,\mathrm dx\le2\pi^{n-1}\int_1^\infty r^{n-p-1}\,\mathrm dr.$$ This is Lebesgue-integrable iff $p>n$ by improper Riemann integration and the dominated convergence Theorem. $\square$
Proof of Lemma 3. This follows directly from $\sigma$-sub-additivity of the integral together with $$\int_{[\mathbf k^1-1,\mathbf k^1[\times\dots\times[\mathbf k^n-1,\mathbf k^n[}\lvert x\rvert^{-p}\ge\lvert\mathbf k\rvert^{-p}$$ for all $\mathbf k\in\mathbb N^{n}$.
[1]: Ivan Netuka, The Change-of-Variables Theorem
for the Lebesgue Integral. ACTA UNIVERSITATIS MATTHIAE BELII, series MATHEMATICS 19 (2011), 37–42. Available online at https://actamath.savbb.sk/pdf/acta1906.pdf.
