Question

[EDIT: I think re-arrange the answer to make the statements clearer].

The answer to the question is no. , for any base field $k$.

Firstremark: if you have a separated scheme , we can characterize smooth affine varieties $X$ which is are étale (or more generally quasi-finite) over an affine variety $A$, then it has to be quasi-affine. This is because by Zariski's Main Theorem, $X$ is an open subscheme of a scheme $\overline{X}$ which is finite over $A$, hence space $\overline{X}$ is affine. So for your question you should restrict to quasi-affine varieties.

Now consider the dimension 1 caseA=\mathbb A^n_k$.Let $C$ be a

A smooth connected and projective curve over a base field $k$ of genus $g>1$. Let $x_0$ be any closed point in $C$ and consider the affine curve variety $X$ equal to $C$ minus $x_0$. Suppose over $X$ k$ is étale over an affine space $A$, then $\Omega_{X/k}$ is isomorphic to $\Omega_{X/k}\simeq \Omega_{A/k}\otimes_{O_A} O_X\simeq O_X$. So $\Omega_{C/k}|_X$ is trivial. This implies that A=\mathbb A^n_k$ if and only if the canonical divisor $K_C$ on sheaf of differentials $C$ \Omega_{X/k}$ is trivial on free of rank $X$, hence linearly equivalent to n$, generated by $d[x_0]$ with n$ closed differential forms $d\deg x_0 = 2g-2$df_1, \ldots, df_n$.

Proof: (1) If $k$ is finite, you choose a point $x_0$ of degree $> 2g-2$, then you get a contradiction.

Over there exists an algebraically closed field, the answer is the same. You take two closed points étale morphism $x_0, x_1$f: X\to A={\rm Spec}k[t_1,\ldots, t_n]$, then you we have an isomorphism $(2g-2)x_1 \Omega_{A/k}\otimes_{O_A} O_X=f^{\star}\Omega_{A/k}\simeq \sim (2g-2)x_0$, so the divisor $[x_1-x_0]$ is a $(2g-2)$-torsion point in Omega_{X/k}$. Then take ${\rm Jac}(C)$, which is impossible because f_i$ equal to the set image of the $[x_1-x_0]$ is infinite while t_i$ in ${\rm Jac}(C)[2g-2]$ is finite.

[EDIT] In fact, there is a necessary and sufficient condition for an affine smooth scheme O(X)$. (2) If $X$ over df_1,\ldots, df_n$ as above exist, we define a field $k$ of dimension morphism $n$ f: X\to A$ to be étale over an affine space $A=\mathbb A^n_k$. On the one hand, if $X$ is étale over $A$, associated to the isomorphism $\Omega_{A/k}\otimes O_X\simeq \Omega_{X/k}$ implies that $\Omega_{X/k}$ is free, generated by moprhism of $df_1, \ldots, df_n$ for some k$-algebras $f_i\in k[t_1,\ldots, t_n]\to O(X)$. On the other hand, if such $f_i$'s exist, then the morphism $X\to A$ defined t_i\mapsto f_i$. Then by hypothesis on the $x\mapsto (f_1(x), \ldots, f_n(x))$ is étale because df_i$'s, the canonical map morphism $\Omega_{A/k}\otimes O_X\to f^{\star}\Omega_{A/k}\to \Omega_{X/k}$ is an isomorphism.

I don't know how to interprete simply this condition on Therefore $\Omega_{X/k}$. Note that f$ is étale.

Remark: it is not enough to say suppose that this sheaf $\Omega_{X/k}$ is free . to insure that $X$ is étale over $A$. For example, if $X$ is an elliptic curve $E$ minus the origin, then $\Omega_{X/k}$ is free , (because $\Omega_{E/k}$ is !), but at least in characteristic $0$, $X$ is not étale over $\mathbb A^1_k$ (if so, it would be finite and étale over $\mathbb A^1$ by considering the extension $E\to \mathbb P^1_k$, but $\mathbb A^1_k$ is simply connected in characteristic $0$). I don't known whether this can happen in positive characteristic. Note that this is already an example (in characteristic 0) of a smooth affine curve which is not étale over $\mathbb A^1$.

Now we construct in any characteristic a smooth affine curve $X$ such that $\Omega_{X/k}$ is not free. By the above, $X$ will not be étale over ${\mathbb A^1}$. Fix a projective smooth connected curve $C$ over $k$ of genus $g>1$. Suppose that for any affine open subset $X$ of $C$, $\Omega_{X/k}$ is free. We want to find a contradiction.

We first reduce to the case $k$ is algebraically closed (for simplicity, this is actually not necessary). I claim that over the algebraic closure $\bar{k}$ of $k$, $\Omega_{X'/\bar{k}}$ is free for any affine open subset $X'$ of $C_{\bar{k}}$. Indeed, $X'$ is defined over a finite extension $K/k$, and the projection $C_{K}\to C$ induces an étale morphism from $X'$ to an affine open subset $X$ of $C$, thus $\Omega_{X'/K}\simeq \Omega_{X/k}\otimes O_{X'}$ is free.

Now we suppose that $k$ is algebraically closed. For any closed point $x\in C$, $\Omega_{X/k}$, where $X=C \setminus {x}$, is free. So $\Omega_{C/k}|_X$ is trivial. This implies that the canonical divisor $K_C$ on $C$ is trivial on $X$, hence linearly equivalent to $(2g-2)[x]$. Fix a point $x_0\in C(k)$. Then the immersion $C\to {\rm Jac}(C)$, $x\mapsto [x-x_0]$, maps $C$ into the $(2g-2)$-torsion part of ${\rm Jac}(C)(k)$ which is finite. Contradicton.

Final remark: if you have a separated scheme $X$ which is étale (or more generally quasi-finite) over any affine scheme $A$, then it has to be quasi-affine. This is because by Zariski's Main Theorem, $X$ is an open subscheme of a scheme $\overline{X}$ which is finite over $A$, hence $\overline{X}$ is affine.