In your comment, you exhibit two important features of “having measure zero”: it is a local property and it is invariant under diffeomorphisms. This is because 1. you are considering a measure continuous with respect to the Lebesgue measure, 2. the ambient space is sigma countable and 3. locally compact

Now let us simplify the problem a bit: we only need to prove that

PROPOSITION. Any algebraic hypersurface of ${\bf C}^N$ is a null set for the Lebesgue measure.

To justify this simplification, observe that $\mu$ is continuous with
respect to the Lebesgue measure, so that it has more null sets
(i.e. sets of measure zero) and it is then enough to show that your
algebraic variety is a null set for the Lebesgue measure. An
algebraic variety is cut out by hypersurfaces, so if any (real)
hypersurface is a null set, any (real) algebraic variety must also be
a null set. The last point to check is that if the proposition holds
it implies the analog statement for real algebraic sets. I will not go in the
details, but the point is that if V is the complex zero locus of a
real equation, you can cover it by acountable set of self-conjugated
measurable compact sets whose union has an arbitrary small measure:
you conclude by observing that the Lebesgue measure of ${\bf R}^N$ can
be recovered from the Lebesgue measure of ${\bf C}^N$ with an
$\epsilon$-based construction similar to yours. (To deal with this
gently, use the cahracterisation of the Lebesgue measure as the
Haar-mesure for the topological group ${\bf R}^N$.)

It is not a very clean exposition, but I nevertheless assume I
convinced you that the PROPOSITION implies that algebraic subvarieties
of ${\bf R}^N$ are null sets for the Lebesgue measure.

Proof of the PROPOSITION. Let $f$ be a complex polynomial on ${\bf
C}^N$ vanishing at some point $p$. Thanks to the Weierstrass
preparation theorem, we can assume that in a coordinate system
centered on $p$, the function $f$ has the form
$$
f(z, \zeta) = z^k + z^{k-1} g_1(\zeta) + \cdots + g_k(\zeta)
$$
where $(z,\eta)$ belongs to some neighbourhood of the origin in ${\bf
C} \times {\bf C}^{N+1}$. Because the (graded) algebra of symmetric
polynomials in $k$ variables spanned by the $k$ elementary symmetric functions
$\sigma_j$ is isomorphic to the (graded) algebra of polynomials in $k$
variables via the map
$$
\Phi: \phi(X_1,\ldots,X_k) \mapsto \phi(\sigma_1, \ldots, \sigma_k)
$$
there is polynomials $\gamma_1, \ldots, \gamma_k$ such that
$$
f(z, \zeta) = \prod_{j = 1}^k (z - \gamma_j(\zeta))
.
$$
In other words we look at the morphism from ${\bf C}^k$ to the set $M$
of monic polynomials of degree $k$ given by
$$
\omega \mapsto \prod_{j = 1}^k (z - \omega_k),
$$
it induces an isomorphism of algebraic varieties ${\bf C}^k /
{\mathfrak{S}_k} \simeq M$ which is precisely $\Phi$. Looking at $f$
as a function of $\zeta$ with values in $M$, we define the $\gamma_j$
(up to permutation!) by composing $f$ with the map $M \to {\bf C}^k /
{\mathfrak{S}_k}$.

This is very good, because we now see that the zero locus of $f$ is
near $p$ the union of the graphs $z = \gamma_j(\eta)$ of $k$
functions. But a graph is diffeomorphic to a hyperplane and is thus a
null set.