This is more of a longish comment explaining why I believe that the answer should be yes, and that a confirmation should be within reach using current technology.
Let $K$ be the field of $n$-th roots of unity, and $\zeta_p$ a primitive $p$-th root of unity for some prime factor $p \mid n$. The question is whether there is a nonsquare unit $\omega \in {\mathcal O}_K^\times$ such that $\omega \equiv 1 - \zeta \bmod 4$.
I can see no reason why such units should not exist, even in the special case $n = p$. But one may have to look for a while before stumbling over an example. The related problem of finding a nonsquare unit $\omega \equiv 1 \bmod 4$, for example, is not solvable for $n = p < 29$ since the corresponding cyclotomic fields have odd class number; ${\mathbb Q}(\zeta_{29})$, on the other hand, has a class group of type $(2,2,2)$ and a good chance of containing such a unit. This can probably be verified by looking only at cyclotomic units, which are known explicitly.
In your case, you should look at products of units of the form $$ \omega = (1+\xi)^{a_1}(1+\xi+\xi^2)^{a_2}(1+\xi+\xi^2+\xi^3)^{a_3} \cdots $$ with not all $a_j$ even, and check whether one of these lies in the residue class $1 - \xi^j \bmod 4$. Using sage or pari, this should actually be doable. Perhaps some linear algebra and the Chinese remainder theorem can be used to speed up the calculations.
Edit. I was so convinced that there would be a solution of the problem for a small $n$ that I did not do what I should have done: the problem in question is equivalent to the congruence $\omega \equiv \alpha^2 (1 - \zeta) \bmod 4$ for some cyclotomic integer $\alpha$ coprime to $2$. I guess Dror's code can easily be adapted to the more general congruence.
