A closed form of $q$-analog finite sum

Question

Let $[n]:=\frac{1-q^n}{1-q}$. I wish to find a closed form of the following q-sum $$I(p):=\sum\limits_{n_1+\cdots+n_p=n\atop n_1,\ldots,n_p\ge 1}\ \ \ \frac{1}{[n_1]\cdots [n_p]}=？$$ For example, we have $$I(2)=\frac{1}{[n]}\sum\limits_{k=1}^{n-1}\frac{1+q^k}{[k]},$$ but how to obtain the general result for $p\ge 3$?