The Lebesgue measure of zero set of a polynomial function is zero
Suppose the theorem is established for polynomials in $n-1$ variables. Let $p$ be a nontrivial polynomial in $n$ variables, say of degree $k \ge 1$ in $x_n$. We can then write $$p(\mathbf{x}, x_n) = \sum_{j=0}^k p_j(\mathbf{x}) x_n^j$$ where $\mathbf{x} = (x_1, \dots, x_{n-1})$ and $p_0, \dots, p_k$ are polynomials in $n-1$ variables, where at least $p_k$ is nontrivial.
Let us note that since $p$ is continuous, the zero set $Z(p)$ is a measurable subset of $\mathbb{R}^n$.
Now if $(\mathbf{x}, x_n)$ is such that $p(\mathbf{x}, x_n) = 0$ then there are two possibilities:
$p_0(\mathbf{x}) = \dots = p_k(\mathbf{x}) = 0$, or
$x_n$ is a root of the (nontrivial) one-variable polynomial $p_{\mathbf{x}}(t) =\sum_{j=0}^k p_j(\mathbf{x}) t^j$.
Let $A,B$ be the subsets of $\mathbb{R}^n$ where these respective conditions hold, so that $Z(p) = A \cup B$.
Use the inductive hypothesis to show $A$ has measure zero.
Use the fundamental theorem of algebra to show that for each fixed $\mathbf{x}$, there are finitely many $t$ such that $(\mathbf{x},t) \in B$. (Indeed, there are at most $k$.) A finite set has measure zero in $\mathbb{R}$. Now apply Fubini's theorem to conclude that $B$ has measure zero. (Note that $B = Z(p) \setminus A$ is measurable.)
As you are going to see, your result is true not only for polynomials, but for analytic functions. Since these have no concept of degree, we shall use the concept of order instead. If
$$f = \sum \limits _{(k_1, \dots, k_n) \in \Bbb N ^n} a_{k_1, \dots, k_n} x_1 ^{k_1} \dots x_n ^{k_n} \ne 0$$
then there exist an $a_{k_1, \dots, k_n} \ne 0$. Let the order of $f$ be $o(f) = \min \{k_1 + \dots + k_n \mid a_{k_1, \dots, k_n} \ne 0\}$. Convince yourself that if $f$ is not constant then $o \left( \dfrac {\partial f} {\partial x_i} \right) = o (f) -1, \ \forall i = 1, \dots, n$.
We shall use induction on $o(f)$ to prove that if $f \ne 0$, then $Z(f)$ has Lebesgue measure $0$.
If $o(f) = 0$, then $f = a_{0, \dots, 0} \in \Bbb R \setminus \{0\}$, so $Z(f) = \emptyset$ and the statement is vacuously true.
Assume $o(f) > 0$ and the statement proven for $\{0, \dots, o(f) - 1\}$.
Let $R(f) = \{x \in Z(f) \mid \nabla f \ne 0 \}$ (the regular points of $f$) and $C(f) = \{x \in Z(f) \mid \nabla f = 0 \}$ (the critical points of $f$). Obviously, $Z(f) = R(f) \cup C(f)$ and $R(f) \cap C(f) = \emptyset$.
If $x \in R(f)$ then there exist some small neighbourhood $U_x \subseteq \Bbb R^n$ of $x$ such that $f \big| _{U_x}$ is a submersion, whence it follows that $R(f) \cap U_x$ is a submanifold of $\Bbb R^n$ of codimension $1$. This shows that $R(f)$ is a submanifold of codimension $1$, so it has Lebesgue measure $0$ (being locally like $\Bbb R ^{n-1}$).
If $x \in C(f)$, then $\dfrac {\partial f} {\partial x_i} (x) = 0, \ \forall i = 1, \dots, n$, so in particular $x \in Z \left( \dfrac {\partial f} {\partial x_1} \right)$, so $C(f) \subseteq Z \left( \dfrac {\partial f} {\partial x_1} \right)$. But as said above, $o \left( \dfrac {\partial f} {\partial x_1} \right) = o(f) - 1$, so by applying the induction hypothesis to $\dfrac {\partial f} {\partial x_1}$ one gets that $Z \left( \dfrac {\partial f} {\partial x_1} \right)$ has Lebesgue measure $0$. Since the Lebesgue measure is regular (i.e. every subset of a null set is itself a null set), it follows that the measure of $C(f)$ is $0$ too.
Since the Lebesgue measure of $Z(f)$ is the sum of the Lebesgue measures of $R(f)$ and $C(f)$, it follows that $Z(f)$ must be a null set.
As was mentioned in a comment above, the result doesn't hold for a general smooth function. Suppose $f(X_1, \dots, X_n)$ is a polynomial, and assume without loss of generality that all $\partial f/\partial X_i\not\equiv 0$. By the constant rank theorem, the result holds off the $Z(\partial f/\partial X_i)$. Now induct on the degree of $f$.