I would try to get rid of the trigonometric functions, and rather rewrite the system as a polynomial system. If $x_j=\cos(\phi_j)$, then $\cos(\phi_j(2\ell+1))=T_{2\ell+1}(x_j)$, where $T_k$ is the $k$-th Chebychev polynomial of degree $k$. So your system of equations is
\begin{equation}
0=1+2\sum_j T_{2\ell+1}(x_j),\;\;l=1,2,\dots,m,
\end{equation}
with the requirement that $0<x_j\le 1$. Such a system can be discussed via Groebner bases, see here how to do that with Sage. For $m=n\le4$ there are only finitely many solutions. Among them pick those which fit your inequalities. For instance, for $m=n=4$, an approximation of a solution seems to be
\begin{align*}
x_1 &= 0.963494595276259\\
x_2 &= 0.852773246361416\\
x_3 &= 0.600336170417163\\
x_4 &= 0.058262327046178
\end{align*}
Of course, you can use this technique also to handle the original case, where $t_j$ need not have length $1$. For instance, for $n=2$ one can show that $m\le4$, and and approximate solution for $m=4$ is $t_1=0.466916296430820 + 0.717248344919154i$, $t_2=0.856453001234213 + 0.445264622297009i$.
One cannot expect simple expressions for the $t_j$. For instance, the absolute values of the $t_j$ are roots of an irreducible (over the rationals) polynomial of degree $60$.
