Solution:
We claim the the minimum possible value of this expression is $n$. Observe that by AM-GM, we have that
$$\begin{aligned} \frac{a_{1}}{1} &+ \frac{a_{2}}{1+a_{1}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}} \\ &=\frac{1+a_{1}}{1}+\frac{1+a_{1}+a_{2}}{1+a_{1}}+\cdots+\frac{1+a_{1}+a_{2}+\cdots+a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}-n \\ &\geq n \cdot \sqrt[n]{\frac{1+a_{1}}{1} \cdot \frac{1+a_{1}+a_{2}}{1+a_{1}} \cdots \frac{1+a_{1}+a_{2}+\cdots+a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}}-n \\ &=n \cdot \sqrt[n]{1+a_{1}+a_{2}+\cdots+a_{n}}-n \\ &=2n-n=n. \end{aligned}$$
Furthermore, equality is achieved when $a_{k}=2^{k-1}$ for each $1 \leq k \leq n$.