Generalizing algebraic numbers to multivariate polynomials

Consider the set of (complex) algebraic numbers $\mathbb{A}$, defined as $\mathbb{A} = \{x \in \mathbb{C} \mid p(x) = 0 \text{ for some $p \in \mathbb{Z}[x]$}\}.$ Suppose we now wanted to generalize this to multivariate polynomials via $$ \mathbb{A}_n = \{(x_1, \dots, x_n) \in \mathbb{C}^n \mid p(x_1,\dots,x_n) = 0 \text{ for some $p \in \mathbb{Z}[x_1, \dots, x_n]$} \}.$$

Right away I noticed that $\mathbb{A}_n \neq \mathbb{A}^n$ for $n \neq 1$, as all pairs of the form $(z,z,0,\dots,0)$ (for complex $z$) are in $\mathbb{A}_n$ since $(z,z,0,\dots,0)$ is a root of the polynomial with integer coefficients $p(x_1,\dots,x_n) = x_1 - x_2.$ In particular when $n \neq 2$ we have $(\pi, \pi) \in \mathbb{A}_2$ but $(\pi, \pi) \not\in \mathbb{A}^2$.

Is this construction meaningful in any way? I know that $\mathbb{A}^n \subseteq \mathbb{A}_n$. Does it hold for $n > 1$ that $\mathbb{A}_n = \mathbb{C}^n$? My guess is no, but I’m not entirely sure how I’d go about proving this (I imagine that a counterexample would be nice, but proving that a set of $n$ numbers satisfies no polynomial with integer coefficients seems a lot harder than proving that any one number doesn’t). What other information, useful or otherwise, can be gathered from the structure of $\mathbb{A}_n$ for $n > 1$?

(I assume you want to require $p\neq 0$ in your definition of $\mathbb{A}_n$ to avoid trivialities.)

This is kind of tautological, but perhaps a more enlightening way of describing your $\mathbb{A}_n$ is that it is the set of $n$-tuples of complex numbers that are not algebraically independent over $\mathbb{Q}$. This makes it immediately clear that $\mathbb{A}_n$ is not all of $\mathbb{C}^n$. Explicitly, you can choose the coordinates one at a time to get a point in the complement: first pick $x_1\in\mathbb{C}$ that is transcendental over $\mathbb{Q}$, then pick $x_2\in\mathbb{C}$ that is transcendental over $\mathbb{Q}(x_1)$, then pick $x_3\in\mathbb{C}$ that is transcendental over $\mathbb{Q}(x_1,x_2)$, and so on. It is always possible to keep choosing more transcendental elements like this since the algebraic closure of $\mathbb{Q}(x_1,\dots,x_k)$ is always countable. (This also illustrates that $\mathbb{A}_n$ is quite "small": for instance, it has Lebesgue measure $0$, since these countable sets on each coordinate have measure $0$. See this answer of mine for some related ideas which also give an alternate way of proving $\mathbb{A}_n$ is not all of $\mathbb{C}^n$.)