It looks to be enough to require this for $\lambda=1$. Indeed, assume that $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$ has all roots on the unit circle, denote the roots $-\theta_i$, $i=1,\ldots,n$. Then $Q(Z)=\prod (Z+\theta_i)$ and $\prod_i \theta_i=1$. There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. It works.