Determinant of the "quantum" version of the group $\mathbb{Z}_n$ Let $[0]_q:=0$ and $[n]_q:=\frac{1-q^n}{1-q}=1+q+\cdots+q^{n-1}$, for $n\geq1$.

Question. Is there a closed formula (with proof) for the determinant of the matrix of $(i,j)$-entries
  $$[i+j\bmod n]_q, \qquad i,j=1,2,\dots,n.$$

Remark. To bring in some context to the problem, this determinant is the specialization 
$$x_{i+j\bmod n}\rightarrow [i+j\bmod n]_q$$ 
in the group determinant of Frobenius for the (finite) additive group $\mathbb{Z}_n$.
EDIT. Sorry, I was meant to write $[0]_q=0$ not $[0]_q=1$.
 A: Let me give a few details on Fedor's calculation. If we swap the columns with index $j$ and $n+1-j$ we get the matrix $M_n(q)$ with $(i,j)$ entry equal to $[1+i-j \mod n]_q$. Since we have swapped $\lfloor \frac{n}{2}\rfloor$ columns, the determinant of your original matrix is equal to
$$(-1)^{\lfloor \frac{n}{2}\rfloor}\det M_n(q)=(-1)^{\frac{n(n-1)}{2}}\det M_n(q).$$
Now, $M_n(q)$ is a circulant matrix so it has eigenvectors $(1,\omega^k,\dots,\omega^{k(n-1)})$, with corresponding eigenvalue $[1]_q+[0]_q\omega^k+[n-1]_q\omega^{2k}+\cdots+[2]_q\omega^{(n-1)k}$, for all $0\le k\le n-1$ where $\omega=\exp(\frac{2\pi i}{n})$.
Next we observe that
$$[1]_q+[0]_q\omega^k+[n-1]_q\omega^{2k}+\cdots+[2]_q\omega^{(n-1)k}=\omega^{k}\left([0]_q+\frac{1+\frac{q}{\omega^k}+\cdots \frac{q^{n-1}}{\omega^{k(n-1)}}}{q-1}\right)=\omega^k\left([0]_q+\frac{q^n-1}{(q-1)(\frac{q}{\omega^k}-1)}\right)$$
when $k\neq 0$. So we can calculate
$$\det M_n(q)=\omega^{\frac{n(n-1)}{2}}\left([0]_q+[1]_q+\cdots+[n-1]_q\right)\prod_{k=1}^{n-1}\frac{[0]_q(q-\omega^k)+\omega^k [n]_q}{q-\omega^k}$$
if you take $[0]_q=0$ this gives 
$$\det M_n(q)=[n]_q^{n-2}\frac{[n]_q-n}{q-1}$$
and if you take $[0]_q=1$ it gives
$$\det M_n(q)=(-q)^{n-1}\left(1+\frac{[n]_q-n}{q-1}\right)\frac{1+(-1)^{n-1}[n-1]_q^n}{[n]_q(1+[n-1]_q)}.$$
A: Replacing $j$ to $n-j$ we see that this determinant equals $(-1)^{n(n-1)/2}$ times circulant corresponding to the polynomial $f(t)=\sum_{i=0}^{n-1}[i]t^i$. We should multiply $f(w)$ when $w$ runs over the roots of $w^n-1$. We have $f(t)\cdot (1-q)=(1-t^n)/(1-t)-(1-t^nq^n)/(1-tq)$. For $t=w$ this simplifies and it remains to multiply $1-qw$ over $w$, that equals $1-q^n$.
