One heuristic is to replace the $n^{th}$ roots of unity by $n$ iid elements $\zeta_1,\dots,\zeta_n$ of the unit circle, drawn uniformly at random. For any sum $\zeta_{i_1} + \dots + \zeta_{i_k}$ of $k$ of these with $i_1 < \dots < i_k$, a standard Fourier analytic calculation (Esseen concentration inequality) shows that if $k \geq 5$ (to make the $k$-fold convolution power of the Fourier transform of the unit circle have absolutely integrable Fourier transform), then $$ {\bf P}( |\zeta_{i_1} + \dots + \zeta_{i_k}| < r ) \ll_k r^2 $$ for $0 < r < 1$. Taking the union bound over all $\binom{n}{k}$ different sums, we see that the probability that $k$ distinct elements of the $\zeta_1,\dots,\zeta_n$ sum to something of magnitude less than $r$ is $O_k( \binom{n}{k} r^2 )$. This shows that with probability $\gg 1$, we should have the lower bound $$ |\zeta_{i_1} + \dots + \zeta_{i_k}| \gg_k \binom{n}{k}^{-1/2} \asymp_k n^{-k/2}$$ for all $i_1 < \dots < i_k$. This is for a fixed $n$. To get a bound that is true with positive probability for all but finitely many $n$, the Borel-Cantelli lemma suggests that we need to get the failure probability down to something like $1/n^{1+\varepsilon}$, and so the lower bound now worsens from $\gg_k n^{-k/2}$ to $\gg_{k,\varepsilon} n^{-(k+1+\varepsilon)/2}$.
To complete the heuristic analysis one should also treat the cases where there are collisions amongst the $i_1,\dots,i_k$, which can be handled by the same methods as long as there are at least $5$ distinct values of the $i_1,\dots,i_k$; the cases of only four or fewer values need to be treated by more direct computations but I think they should give lower order contributions in the $k \geq 5$ regime.