This is about the univariate case ($n = 1$) only, but way too long for a comment.

*Step 1: notation.* Let $u$ be a continuous function of bounded variation, let $\mu = Du$ be the corresponding Radon measure, let $|\mu|$ denote the variation measure of $\mu$, and let $U$ be the distribution function of $|\mu|$:
$$ U(x) = \begin{cases} |\mu|([0,x)) & \text{if } x \geqslant 0 \\ |\mu|([x, 0)) & \text{if } x < 0 \end{cases} $$
(so that $|\mu| = D U$). Let $S$ be the support of $\mu$, and let $S'$ be equal to $S$ right endpoints of every connected component of $S$ removed. Then $|\mu|$ assigns no measure to the complement of $S'$ (because it assigns no measure both to the complement of $S$ and to the countable set $S \setminus S'$), and $U$ is a one-to-one mapping of $S'$ onto some interval $I$. The notation would have been much simpler if we assumed that $S$ is an interval, but let us deal with the general case.

Define a measure $\nu$ on $I$ by the formula
$$ \nu(B) = \mu(U^{-1}(B)) = \mu(U^{-1}(B) \cap S') . $$
Since $U : S' \to I$ is a bijection and $\mu(A) = \mu(A \cap S') = \nu(U(A))$, we have
$$ \nu_\pm(B) = \mu_\pm(U^{-1}(B)) . $$
Therefore,
$$ |\nu|(B) = |\mu|(U^{-1}(B)) = |B| $$
is the Lebesgue measure of $B$ (as long as $B$ is contained in $I$, of course). It follows that $\nu$ is absolutely continuous on $I$, with density function equal to $\pm 1$ almost everywhere.

Let $w(y) = u(U^{-1}(y))$, where $U^{-1}(y)$ is the unique point $x$ in $S'$ such that $U(x) = y$ (note, however, that since $v$ is constant on every connected component of the complement of $S$, any other choice of $x$ would work equally well). It is easy to check that $w$ is the distribution function of $\nu$ (that is, $\nu = D w$), so that $w'$ exists almost everywhere, and $|w'| = 1$ almost everywhere.

Clearly, for any $p$ we have
$$|\mu|(u^{-1}(\{p\})) = |\mu|(u^{-1}(\{p\}) \cap S') = |\nu|(U(u^{-1}(\{p\}))) = |\nu|(w^{-1}(\{p\})) = |w^{-1}(\{p\})|.$$
Note that $w' = 0$ almost everywhere on $w^{-1}(\{p\})$ (indeed: if $w'(y)$ exists for some non-isolated point $y$ in $w^{-1}(\{p\})$, then $w'(y) = 0$). However, $|w'| = 1$ almost everywhere. This means that $w^{-1}(\{p\})$ has zero Lebesgue measure, and consequently
$$ |\mu|(u^{-1}(\{p\})) = 0 ,$$
as desired.

I have no experience with multivariate functions of bounded variation, so I will stop here. Let me remark, however, that the multivariate case might follow from the univariate one if, for example, there is a bounded variation counterpart of the "absolutely continuous on lines" (ACL) characterisation of weakly differentiable functions.

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