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][1] for the (finite) additive group $\mathbb{Z}_n$.

**EDIT.** Sorry, I was meant to write $[0]_q=0$.

[1]: https://mathoverflow.net/questions/251971/converse-of-frobenius