I assume that $n=p$ be a prime and $p>3$: indeed, if $p=3$ then your statement is false because $\xi^2-\xi+1=0$. Your polynomial is $$ \chi(\lambda)=\lambda^2+(\xi-1)^2\lambda+\lambda\xi+\xi^2\in\mathbb{Z}[\xi]. $$ where $\xi$ is a primitive $2p$-th root of unity. Let $\zeta$ be a root of $\chi$ and assume it is a root of unity: hence, as Sebastian Schoennenbeck observed, either $\zeta\in\mathbb{Q}(\xi)=\mathbb{Q}(\mu_{2p})$ or it lies in the quadratic extension $\mathbb{Q}_{4p}$. The latter is not the case since the minimal polynomial of $\zeta$ over $\mathbb{Q}(\mu_{2p})$ would be $\lambda^2-\xi$, therefore $\zeta\in \langle\xi\rangle$, the cyclic group of order $2p$ generated by $\xi$ inside $\mathbb{Z}[\xi]^\times$. To show that this is not the case, observe that all roots of unity in $\mathbb{Z}[\xi]$ reduce to $\pm 1$ modulo the maximal ideal $\mathfrak{p}\subseteq \mathbb{Z}[\xi]$ which lie above $p$ and contains (is in fact generated by) $\xi-1$ (see Proposition 2.8 of Washingtons's Introduction to Cyclotomic Fields). If we reduce the polynomial  $\mod\mathfrak{p}$ we thus find $$ \bar{\chi}(\lambda)=\lambda^2\pm \lambda+1 $$ and neither $1$ nor $-1$ are roots of either of these polynomials unless $p=3$ which we excluded. Hence, no roots of unity can be a root of $\chi$.

The argument above extends immediately to the case when $n=p^e$ is a prime power but does not extend to the case where $n=p^aq^b$ is composite because $\xi-1$ becomes a unit.

I assume that $n=p$ be a prime and $p>3$: indeed, if $p=3$ then your statement is false because $\xi^2-\xi+1=0$. Your polynomial is $$ \chi(\lambda)=\lambda^2+(\xi-1)^2\lambda+\lambda\xi+\xi^2\in\mathbb{Z}[\xi] $$ where $\xi$ is a primitive $2p$-th root of unity. Let $\zeta$ be a root of $\chi$ and assume it is a root of unity: hence, as Sebastian Schoennenbeck observed, either $\zeta\in\mathbb{Q}(\xi)=\mathbb{Q}(\mu_{2p})$ or it lies in the quadratic extension $\mathbb{Q}(\mu_{4p})$. The latter is not the case since the minimal polynomial of $\zeta$ over $\mathbb{Q}(\mu_{2p})$ would be $\lambda^2-\xi$, therefore $\zeta\in \langle\xi\rangle$, the cyclic group of order $2p$ generated by $\xi$ inside $\mathbb{Z}[\xi]^\times$. To show that this is not the case, observe that all roots of unity in $\mathbb{Z}[\xi]$ reduce to $\pm 1$ modulo the maximal ideal $\mathfrak{p}\subseteq \mathbb{Z}[\xi]$ which lies above $p$ and contains (is in fact generated by) $\xi-1$ (see Proposition 2.8 of Washingtons's Introduction to Cyclotomic Fields). If we reduce the polynomial$\mod\mathfrak{p}$ we thus find $$ \bar{\chi}(\lambda)=\lambda^2\pm \lambda+1 $$ and neither $1$ nor $-1$ are roots of either of these polynomials unless $p=3$ which we excluded. Hence, no roots of unity can be a root of $\chi$.

The argument above extends immediately to the case when $n=p^e$ is a prime power but does not extend to the case where $n=p^aq^b$ is composite because $\xi-1$ becomes a unit.