What follows is not mine. It was posted by Scaroth, and then deleted by Scaroth, with the explanation, "Given the subsequent comments of the OP, the indication was that this answer was not even read, so I will delete it." However, the answer was read by at least 9 people who gave it upvotes. Whether or not it helped OP, several people found it helpful. I have no way to contact Scaroth to ask him/her to reconsider, so I will simply repost it.

$\newcommand\p{\mathfrak{p}}$
$\newcommand\Q{\mathbf{Q}}$

Denote the cylotomic field by $F$.
Since complex conjugation commutes with every automorphism of $G = \mathrm{Gal}(F/\mathbf{Q})$,
it follows that complex conjugation preserves absolute values in $F$. Hence you are asking exactly
for a classification of so-called * $n^2$-Weil numbers*. (By definition,
$m$-Weil numbers are algebraic integers which have absolute value $\sqrt{m}$ for every complex embedding.)
It is generally thought
that if you * fix * an integer $m$, then there are only finitely many $m$-Weil numbers
up to roots of unity in * all* cyclotomic
fields simultaneously, but this is not yet known. On the other hand, if you fix $F$ and increase $m$,
such numbers are easy to come by. Indeed, it's easy to find such $\beta$ so that
the Galois group $G$ of $F$ acts faithfully on the conjugates of not only
the element $\beta$, but the * ideal * $(\beta)$ as well (which is not the case
for $(\zeta \alpha)$ for a root of unity $\zeta$ whenever $\mathbf{Q}(\alpha) \ne F$.)
The answer to your question, therefore, is "there are a lot of them."
There's no easy classification
of Weil numbers, however,
since otherwise the conjecture
mentioned above would be known. If there's a more precise property of the set of numbers that you are interested in, please ask a follow up question.

Recall that, in a cylcotomic field (more generally, a CM field), the real units have finite index
inside $\mathcal{O}^{\times}_F$. It follows that the group of units of the form $\epsilon \cdot \overline{\epsilon}$
has finite index, and hence that there exists a integer $N$ such that:

For every (totally) real unit $\eta$, the unit $\eta^N$
is of the form $\epsilon \cdot \overline{\epsilon}$ for some unit $\epsilon$.

Let us now construct some Weil numbers.
Suppose that the prime $p$ splits completely in $F$.
There are $[F:\mathbf{Q}] = 2d$ such primes, and thus $2^d$ ways to choose a set $S$ of $d$
of primes which includes exactly one from each pair $\{\p,\overline{\p}}$.
Let
$$I = \prod_{S} \p.$$
Then $I \overline{I} = (p)$. Let $h$ denote the class number of $F$, and write
$I^h = (\alpha)$. Then
$$\alpha \overline{\alpha} = p^h \eta$$
for some totally real unit $\eta$. It follows that
$$\alpha^{N} \overline{\alpha}^N = p^{hN} \epsilon \overline{\epsilon},$$
and hence that
$\beta:=\alpha^N/\epsilon$ has absolute value $p^{hN}$. Moreover, from the factorization of $\beta$,
we see that the ideal $(\beta)$ is only fixed by $H \subset \mathrm{Gal}(F/\mathbf{Q})$
which fix the set $S$. In particular, if $|G| > 4$, one can choose $S$ such that the stabilizer of $\beta$
is trivial and thus $F = \Q(\beta)$.

There are many variations on the above argument. For example, one can probably find $p$-Weil
numbers for prime $p$ by finding primes $p$ which split completely in some ray class field of conductor
$M \infty$ (designed so that totally positive units which are $1 \mod M$ are of the
form $\epsilon \cdot \overline{\epsilon}$).

(Given the subsequent comments of the OP, the indication was that this answer was not even read, so I will delete it.)