The answer to your question is yes when $n$ is squarefree.
Here is a general result. When $L/K$ is Galois with Galois group $G$, $L = K(\alpha)$, and $H$ is a subgroup of $G$, then we can ask whether $L^H = K(\alpha_H)$ where $\alpha_H = \sum_{\sigma \in H} \sigma(\alpha) = {\rm Tr}_{L/L^H}(\alpha)$. Easily $\alpha_H \in L^H$, so $K(\alpha_H)$ is a subfield of $L^H$, but it is not at all the case in general that $K(\alpha_H)$ is all of $L^H$. However, we do have $L^H = K(\alpha_H)$ for all subgroups $H$ when $\{\sigma(\alpha) :\sigma \in G\}$, the set of all $K$-conjugates of $\alpha$, is a $K$-basis of $L$. That is called a normal basis of $L/K$. See Theorem 3.8 here.
So the answer to your question is yes when $\zeta_n$ and its $\mathbf Q$-conjugates form a $\mathbf Q$-basis of $\mathbf Q(\zeta_n)$, and that condition is known to happen if and only if $n$ is squarefree. See my answer here.
There is nothing special about using powers of $2 \bmod n$: when $n$ is squarefree and $H$ is any subgroup of $(\mathbf Z/n\mathbf Z)^\times$, the field $\mathbf Q(\sum_{a \in H} \zeta_n^a)$ has degree $\varphi(n)/|H|$ over $\mathbf Q$ by the results I described above.
When $n$ is not squarefree there are counterexamples. Suppose $2 \bmod n$ generates $(\mathbf Z/n\mathbf Z)^\times$. Then your sum is $\sum_{k=1}^{\varphi(n)} \zeta^{2^k} = \sum_{(a,n) = 1}\zeta_n^a = {\rm Tr}_{\mathbf Q(\zeta_n)/\mathbf Q}(\zeta_n)$, which turns out to be $\mu(n)$ for all $n$. So when $n$ is not squarefree, $\mu(n) = 0$ and thus the powers $\{\zeta_n^a : (a,n) = 1\}$ sum to $0$. So those powers are not linearly independent over $\mathbf Q$. And in your case the sum you are looking at is $0$. This happens when $n = 9$: $2 \bmod 9$ generates all units mod $9$ and $\sum_{k=1}^6 \zeta_9^{2^k} = 0$.
