Since 
$$
\frac{1}{1-q^{i_{1}+i_{2}}}\frac{1}{1-q^{i_{2}+i_{3}}}\ldots\frac{1}{1-q^{i_{n}+i_{1}}}>1,
$$
we find
\begin{align*}
f_{n}(q) & >\sum_{i_{1}\ldots i_{n}\geq1}q^{i_1+\ldots+i_n}\\
 & =\left(\sum_{i\geq1}q^{i}\right)^{n}\\
 & =\left(\frac{1}{1-q}\right)^{n}\to\infty.
\end{align*}
Additionally, When $q<\frac{1}{2}, $$\frac{\sqrt{q}}{1-q^{2}}<1,$ and for all $n\geq1$
\begin{align*}
\left(1-q^{2}\right)^{n}+\left(q^{2}\right)^{n} & \leq1\\
\Leftrightarrow\left(1-q^{2}\right)^{n} & \leq\left(1-q^{2n}\right)\\
\Leftrightarrow q^{n}\left(1-q^{2}\right)^{n} & \leq q^{n}\left(1-q^{2n}\right)\\
\Leftrightarrow\frac{q^{n}}{1-q^{2n}} & \leq\left(\frac{q}{1-q^{2}}\right)^{n}
\end{align*}
So 
\begin{align*}
f_{n}(q)=\sum_{i_{1}\ldots i_{n}\geq1}\frac{\left(\sqrt{q}\right)^{i_{1}+i_{2}}}{1-q^{i_{1}+i_{2}}}\frac{\left(\sqrt{q}\right)^{i_{2}+i_{3}}}{1-q^{i_{2}+i_{3}}}\ldots\frac{\left(\sqrt{q}\right)^{i_{n}+i_{1}}}{1-q^{i_{n}+i_{1}}} & \leq\sum_{i_{1}\ldots i_{n}\geq1}\left(\frac{\sqrt{q}}{1-q^{2}}\right)^{2\left(i_{i}+\ldots+i_{n}\right)}\\
 & =\left(\sum_{i=1}^{\infty}\left(\frac{q}{\left(1-q^{2}\right)^{2}}\right)^{i}\right)^{n}\\
 & =\left(\frac{\left(1-q^{2}\right)^{2}}{\left(1-q^{2}\right)^{2}-q}\right)^{n}
\end{align*}
Therefore, when $q\in(0,\frac 12)$ $$\left(\frac{1}{1-q}\right)^{n}\leq f_{n}(q)\leq\left(\frac{\left(1-q^{2}\right)^{2}}{\left(1-q^{2}\right)^{2}-q}\right)^{n}.$$