A bunch of points:
$\def\Spec{\mathrm{Spec}\ }$

• Let $A$ be a cluster algebra over a field $k$, let $(x_1, \ldots, x_n)$ be a cluster and let $L$ be the Laurent polynomial ring $k[x_1^{\pm}, \ldots, x_n^{\pm}]$. I imaging your intended question is whether the map $\Spec L \to \Spec A$ is an open immersion. (You ask about a map $\mathbb{A}^n \to X$, but there isn't a natural such map; $\Spec L$ is a torus, not affine space.)

The answer is yes. The question is equivalent to asking whether $L$ is a localization of $A$. I claim that $L = A[x_1^{-1}, \ldots, x_n^{-1}]$. Proof: On the one hand, $A \subset L$ (by the Laurent phenomenon) and $x_1^{-1}$, ..., $x_n^{-1} \in L$ (obviously), so $A[x_1^{-1}, \ldots, x_n^{-1}] \subseteq L$. On the other hand, $L$ is generated by the $x_j$ and their reciprocals, and these are all in $A[x_1^{-1}, \ldots, x_n^{-1}]$, so $L \subseteq A[x_1^{-1}, \ldots, x_n^{-1}]$. $\square$.

• $\Spec A$ is not generally the union of cluster tori. This is true even in the simplest case: The extended exchange matrix $\left( \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right)$ gives the cluster algebra
$$\mathbb{C}[x,x', y^{\pm 1}]/x x'-y-1.$$
The two tori are $xy \neq 0$ and $x' y \neq 0$. The point $(x,x',y) = (0,0,-1)$ is in neither torus.

• Some sources define "the cluster variety" as the (quasi-affine) union of cluster tori. If you take that as the definition, it is obviously smooth.

• But I imagine what you care about is whether $\Spec A$ is smooth (or possibly $\Spec U$, where $U$ is the upper cluster algebra. No, that doesn't have to be smooth. (The singular points are not in any of the cluster tori, of course.) I think the simplest example is the Markov clsuter algebra, with $B$-matrix $\left( \begin{smallmatrix} 0 & 2 & -2 \\ -2 & 0 & 2 \\ 2 & -2 & 0 \end{smallmatrix} \right)$. This ring is not finitely generated, and there is a maximal ideal generated by all cluster variables where the Zariski tangent space is infinite dimensional. If you like at the upper cluster algebra instead, it is $k[\lambda, x_1, x_2, x_3]/x_1 x_2 x_3 \lambda - x_1^2- x_2^2 - x_3^3$, which is singular along the line $x_1 = x_2 = x_3 = 0$.

For another example, which doesn't have the $A$ versus $U$ issue, look at the $A_3$ cluster algebra with no frozen variables. From Corollary 1.17 in Cluster Algebras III, this is generated by $(x_1, x_2, x_3, x'_1, x'_2, x'_3)$ module the relations
$$x_1 x'_1 = x_2+1,\ x_2 x'_2 = x_1 + x_3,\ x_3 x'_3 = x_2+1.$$

Look at the point $(x_1, x_2, x_3 , x'_1, x'_2, x'_3) = (0, -1, 0, 0,0,0)$. The Jacobian matrix of these $3$ equations with respect to the $6$ variables is
$$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix}$$
which has rank two, so this is a singular point.

• We do have Theorem 7.7 of Greg Muller's Locally Acyclic Cluster Algebras -- if the cluster algebra is locally acyclic, the $B$-matrix has full rank and our ground field has characteristic zero, then $\Spec A$ is smooth.

See Muller 1 and Benito-Muller-Rajchgot-Smith 2 for more on singularities of cluster varieties.