I was thinking, you can send $\sum f(g)g\in\mathbb{Q}[\mathbb{Z}/n\mathbb{Z}]$ to $(f(g))_{g\in G}\in \mathbb{Q}\times\cdots\times\mathbb{Q}$. This is surjective and has zero kernel, so this map should be an isomorphism of $\mathbb{Q}$-algebras.
Should this argument also hold for $\mathbb{C}[\mathbb{Z}/n\mathbb{Z}]$?
On
Greg Martin has already pointed out the flaw. It is easy to see that $\mathbf{Q}[\mathbf{Z}/n\mathbf{Z}]$ is isomorphic as a $\mathbf{Q}$-algebra to the quotient of a polynomial algebra, viz., $\mathbf{Q}[X]/\langle X^n-1\rangle$. And this by the famous Gauss factorization, and Chinese Remainder Theorem, is isomorphic to the product of cyclotomic fields $\prod_{d\mid n}\mathbf{Q}[\zeta_d]$ where $\zeta_d$ is a primitive $d$-th root of $1$.
The premise of your question is wrong: the map you describe does not respect multiplication and is thus not an algebra homomorphism (nor even a ring homomorphism). For instance, look at the first example described here, and track through how your map does not preserve the product of the two given elements. Your map is a $\Bbb Q$-vector space isomorphism, but algebras have more structure than vector spaces.