Fix a norm $|\cdot| : \mathbb{R}^{d} \to [0,\infty)$ on $\mathbb{R}^{d}$. Recall that $\|\cdot\|$ is equivalent to the Euclidean norm $\|\cdot\|$, that is, there is a $C > 1$ such that
\begin{equation*}
C^{-1} \|v\| \leq |v| \leq C\|v\| \quad \text{if} \, \, v \in \mathbb{R}^{d}.
\end{equation*}
Hence $v \mapsto |v|$ is Lipschitz as $||w| - |v|| \leq |w - v| \leq C\|w - v\|$. By the co-area formula,
\begin{equation*}
\int_{\{y \in \mathbb{R}^{d} \, \mid \, |y| \leq 1\}} \|D|\cdot|(x)\| \, dx = \int_{0}^{1} P(\{y \in \mathbb{R}^{d} \, \mid \, |y| = r\}) \, dr
\end{equation*}
Hence there is an $r \in (0,1)$ such that $P(\{y \in \mathbb{R}^{d} \, \mid \, |y| = r\}) < \infty$. At the same time, for any $s > 0$, we have
\begin{equation*}
P(\{y \in \mathbb{R}^{d} \, \mid \, |y| = r\}) = P(r s^{-1} \{y \in \mathbb{R}^{d} \, \mid \, |y| = s\}) = (r s^{-1})^{d-1} P(\{y \in \mathbb{R}^{d} \, \mid \, |y| = 1\}).
\end{equation*}
Thus, the $|\cdot|$-ball centered at the origin has finite perimeter, no matter the radius. Perimeter is translation-invariant, though, so any ball has finite perimeter, no matter the center or the radius.
Another proof I can think of uses the fact that if $C_{1}, C_{2}$ are convex sets and $C_{1} \subseteq C_{2}$, then $P(C_{1}) \leq P(C_{2})$: see here, for instance. I suspect there are other easier proofs.
Edit: It's worth remarking that the argument above shows any convex set $C$ has locally finite perimeter.
First, note that if $C$ has empty interior, then it is contained in an affine hyperplane. In this case, $C$ has Lebesgue measure zero so it trivially has finite perimeter. We assume henceforth that $C$ has nonempty interior.
Note that if $C$ is convex, then $\text{int}(C)$ is also convex. Let us show, then, that open convex sets have locally finite perimeter.
One concludes by arguing that $C \setminus \text{int}(C)$ has zero Lebesgue measure.
Fix $x_{0} \in C$ and $R > 0$ arbitrary. Define the Minkowski gauge $\psi_{R}$ by
\begin{equation*}
\psi_{R}(y) = \inf \left\{ \alpha \, \mid \, \alpha^{-1} y \in (C - x_{0}) \cap B(0,R) \right\}.
\end{equation*}
It is well-known (see, for example, Brezis's functional analysis textbook) that $\psi_{R}$ is a Finsler norm: it is convex and one-homogeneous. (It satisfies all the hypotheses of a norm, except for symmetry.) In particular, as in the case of a genuine norm, $\psi_{R}$ is Lipschitz.
Arguing as above, we see that $\{y \in \mathbb{R}^{d} \, \mid \, \psi_{R}(y) = s\}$ has finite perimeter for each $s \in (0,\infty)$. Translating (and setting $s = 1$), this implies that $C \cap B(x_{0},R)$ has finite perimeter. Given that this is true for all $R > 0$, we conclude $C$ has finite perimeter.
(Other proofs might be shorter, but recently I find working with gauges makes proving things easier.)
