1. According to the properties of logarithms, after transformations we get

$$
\begin{aligned}
& \log _{2}\left(b_{2} b_{3} \ldots b_{n}\right)=\log _{2} b_{1}^{2} \\
& b_{2} b_{3} \ldots b_{n}=b_{1}^{2}
\end{aligned}
$$

Using the formula for the general term of a geometric progression, we get

$$
b_{2}=b_{1} q, b_{3}=b_{1} q^{2}, \ldots, b_{n}=b_{1} q^{n-1}
$$

and using these relations and the formula for the sum of an arithmetic progression

$$
1+2+\ldots+(n-1)=\frac{n(n-1)}{2}
$$

we transform the obtained equation to

$$
b_{1}^{n-1} q^{1+2+\ldots+(n-1)}=b_{1}^{2} \Rightarrow b_{1}^{n-3} q^{n(n-1) / 2}=1
$$

Taking the logarithm with base $q$, we find

$$
\begin{aligned}
& (n-3) \log _{q} b_{1}+\frac{n(n-1)}{2}=0 \\
& \frac{n-3}{2} \log _{q} b_{1}^{2}+\frac{n(n-1)}{2}=0 \\
& \log _{q} b_{1}^{2}=-\frac{n(n-1)}{n-3}
\end{aligned}
$$

Since

$$
\frac{n(n-1)}{n-3}=\frac{n(n-3)+2(n-3)+6}{n-3}=n+2+\frac{6}{n-3}
$$

then

$$
\log _{q} b_{1}^{2}=-n-2-\frac{6}{n-3} .
$$

According to the problem's condition, the quantity

$$
\frac{6}{n-3}
$$

must be an integer. This is possible only for

$$
n \in\{2 ; 4 ; 5 ; 6 ; 9\}
$$

Substituting the specified values into the formula obtained above, we find all possible values of the desired quantity

$$
\begin{aligned}
& n=2 \Rightarrow \log _{q} b_{1}^{2}=2 \\
& n=4 \Rightarrow \log _{q} b_{1}^{2}=-12 \\
& n=5 \Rightarrow \log _{q} b_{1}^{2}=-10 \\
& n=6 \Rightarrow \log _{q} b_{1}^{2}=-10 \\
& n=9 \Rightarrow \log _{q} b_{1}^{2}=-12
\end{aligned}
$$

From this, it is clear that the smallest value of the desired quantity is -12 and it is achieved for two values: $n=4$ and $n=9$.
Answer: -12 when $n=4$ and $n=9$.