Let $A$ be any finite étale $\mathbb{Q}$-algebra (i.e. it's isomorphic to a finite direct product of number fields).
Let $S$ be the set of all $\mathbb{Q}$-algebra homomorphisms $A \to \bar{\mathbb{Q}}.$
I'm trying to show that the natural map $$\Phi:A\otimes_\mathbb{Q}\bar{\mathbb{Q}}\longrightarrow \text{Hom}_{\,\text{Set}}(S,\bar{\mathbb{Q}})$$ given by $$\Phi:a\otimes x \longmapsto \big(s \mapsto s(a)x\big)$$ is an isomorphism of $\bar{\mathbb{Q}}$-vector spaces.
I have been able to show that $S$ is a finite set of size $\dim_\mathbb{Q}A.$
Hence $\dim_{\bar{\mathbb{Q}}}(A\otimes_\mathbb{Q}\bar{\mathbb{Q}})=\dim_{\mathbb{Q}}(\text{Hom}_{\,\text{Set}}(S,\bar{\mathbb{Q}})).$
It therefore suffices to either that $\Phi$ is injective or $\Phi$ is surjective.
I feel that surjectivity will be easier to show but I'm stuck and would really appreciate some help!
As you correctly conjectured, it suffices to prove that the map $\Phi$ is surjective.
If it were not the case, then there would exist a hyperplane $H\subset Map(S,\overline {\mathbb Q})$ containing the image of $\Phi$ .
That hyperplane would be defined by a family $\lambda =(\lambda_s)$ of numbers $\lambda_s\in \bar{\mathbb{Q}}$ and would be of the form $$H_\lambda =\{\alpha:S\to \bar Q\vert \sum_{s\in S} \lambda_s\alpha(s)=0\} $$ Since $\Phi(A)\subset \operatorname {Im}(\Phi)=\Phi(A\otimes_\mathbb{Q}\bar{\mathbb{Q}})\subset H_\lambda$ the equality $\sum_{s\in S}\lambda_s s(a)=0$ would obtain for all $a\in A$.
But that would contradict Dedekind's theorem on independence of homomorphisms, namely Bourbaki, Algebra II, Chapter v, Theorem 1, page 27.
.