For $n$ prime, there seem to be standard ways of determining the subfields. This question gives such a method. However, this method does not necessarily work for $n$ not prime, and the method that I have employed seems very ad hoc. Essentially, my method has just been to determine the Galois group and find its subgroups, look to see if any roots are fixed or cycle through each other, and then trying adding or multiplying those roots and checking if the minimal polynomials are the same as the degree of the extensions.
An Example
For example, consider $\zeta = \zeta_{18}$. I know that $\Phi_{18} = x^6-x^3+1$, so $\Bbb{Q}(\zeta)$ is of degree $6$ over $\Bbb{Q}$. Further, if we let $\sigma(\zeta) = \zeta^5$, we can completely classify the Galois group of $\Bbb{Q}(\zeta)$: $$\operatorname{Gal}(\Bbb{Q}(\zeta)/\Bbb{Q})= \{e,\sigma, \sigma^2,\sigma^3,\sigma^4,\sigma^5\} \simeq \Bbb{Z}_6.$$
Then I know that by the fundamental theorem, there should be subfields corresponding to each of the following subgroups: \begin{align*} H_1 &= \{e\}, \\ H_2 &= \langle \sigma_3 \rangle = \{e,\sigma^3\}, \\ H_3 &= \langle \sigma^2\rangle = \{e,\sigma^2, \sigma^4\},\\ H_4 &= \operatorname{Gal}(\Bbb{Q}(\zeta)/\Bbb{Q}). \end{align*}
So obviously the fixed field of $H_1$ is $\Bbb{Q}(\zeta)$ and the fixed field of $H_4$ is $\Bbb{Q}$, so I just need to figure out the fixed fields of $H_2$ and $H_3$.
So I noticed that $\sigma^3$ takes $\zeta \mapsto \zeta^{-1} \mapsto \zeta$, $\zeta^5\mapsto \zeta^{-5} \mapsto \zeta^5$, and $\zeta^7\mapsto \zeta^{-7} \mapsto \zeta^7$, so clearly $H_2$ fixes $\zeta+\zeta^{-1}$, $\zeta^5+\zeta^{-5}$, and $\zeta^7+\zeta^{-7}$. Here's where I encounter my first question. By looking on Wolfram I know that each of these values have the same minimal polynomial, namely $x^3-3x-1$, and this is the correct degree for the extension we are looking for. Does this then mean that the fixed field MUST be $\Bbb{Q}(\zeta + \zeta^{-1})$? (Side question: to figure out the above, I relied on Wolfram to tell me the minimal polynomial. Is there a way of figuring out the minimal polynomial of such ugly numbers by hand?)
If we now consider $H_3$, I noticed that both $\sigma^2$ and $\sigma^4$ cycle $\zeta, \zeta^7,\zeta^{-5}$ to each other, and they cycle $\zeta^{-1},\zeta^5,\zeta^{-7}$ to each other. So my first guess was that the fixed field must be $\Bbb{Q}(\zeta+\zeta^7+\zeta^{-5})$, but checking Wolfram tells me that $\zeta+\zeta^7+\zeta^{-5} = 0,$ so clearly this doesn't work. (Side question: is there a way I would have been able to tell this was equal to $0$ without using Wolfram?) So I then guessed that maybe it was $\zeta\zeta^7\zeta^{-5} = e^{\frac{\pi i}{3}}$. This has a minimal polynomial of degree 2, which matches the degree we want for the extension over $\Bbb{Q}$. So again, does this mean that this is the extension we want?
Summary
Clearly the method I have employed to solve this type of problem is quite ad hoc. So is there a better way of determining the subfields of such an extension other than using the correspondence and just kind of guessing and checking?
Actually the theory of Gaussian periods (which Gauss developped in his study of the construction of regular $n$-gons by ruler and compass) allows to describe explicitly the subfields of the cyclotomic field $K_n=\mathbf Q (\zeta_n)$ when $n$ is square free. If $n$ has a factor $p^2$, there is a conjectural characterization (see precisely below). Here is brief summary:
Fix $n$ and drop the index $n$ in the notations. Write $\mathcal G =\mathrm{Gal}(K/\mathbf Q) \cong (\mathbf Z/n\mathbf Z)^*$. For any subgroup $G \le \mathcal G$, for any $a \in (\mathbf Z/n\mathbf Z)^* $, define the Gaussian period $S(G, a):=\sum g({\zeta}^{a})$ for $g\in G$ (so it's a special trace value, and in particular $S(G,1)\in K^{G}$). The Gaussian periods behave well under Galois action, in particular, if $H \le G$, then $S(G, a)= \sum s(S(H,a))$ for $s$ running through $G/H$.
Theorem: The periods $S(G,b)$ for all $b\in \mathcal G/G$ form an integral basis of the ring of integers of $K^{G}$ iff $n$ is square-free. A lemma in the course of the proof states that if $n$ is square-free, then $K^{G}=\mathbf Q(S(G,1))$, which answers your question in this case.
Conjecture (based on numerical examples): $K^{G}=\mathbf Q(S(G,1))$ iff $G$ contains no subgroup of the form $\left<1+ \frac n p\right>$ for any prime $p$ s.t. $p^2\mid n$.