UPDATE. Formulae are generalized to include the case $k=1$.
Let us assume that $a$ is fixed, and define: $$ I_n^{k,l}(b,c) := \int\limits_{[1,a]^n} \frac {\chi(\{ b \le x_1 \cdots x_n \le c \})\log(x_1 x_2 \cdots x_n)^l} {( x_1 x_2 \cdots x_n)^k} \,dx_1 \cdots dx_n. $$ Hence, the integral in question equals $I_n^{k,0}(b,c)$.
We first consider the case $b\leq\frac{c}a$, while the other case is considered similarly. We will need the formula: $$\int \frac{\log(y)^l}{y^k}dy = \begin{cases} \frac{\log(y)^{l+1}}{l+1}, & \text{if }k=1;\\ -\sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \frac{\log(y)^i}{y^{k-1}} & \text{if }k\ne 1. \end{cases}$$
Making a substitution $x_n := \frac{t}{x_1\cdots x_{n-1}}$, we get $$I_n^{k,l}(b,c) = \int_1^a dx_1 \cdots \int_1^a dx_{n-1} \int_{\max\{b,x_1\cdots x_{n-1}\}}^{\min\{c,ax_1\cdots x_{n-1}\}} \frac{\log(t)^l dt}{t^k x_1\cdots x_{n-1}}.$$
If $k\ne 1$, depending on the value of $x_1\cdots x_{n-1}$ the last integral breaks into 3 cases:
- If $x_1\cdots x_{n-1}\leq b$, then the integral over $t$ equals $$-\sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \frac{\log(ax_1\cdots x_{n-1})^i}{a^{k-1}(x_1\cdots x_{n-1})^k} + \sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \frac{\log(b)^i}{b^{k-1}x_1\cdots x_{n-1}}$$
- If $b\leq x_1\cdots x_{n-1}\leq \frac{c}{a}$, then the integral over $t$ equals $$-\sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \frac{\log(ax_1\cdots x_{n-1})^i}{a^{k-1}(x_1\cdots x_{n-1})^k} + \sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \frac{\log(x_1\cdots x_{n-1})^i}{(x_1\cdots x_{n-1})^k}.$$
- If $\frac{c}{a}\leq x_1\cdots x_{n-1}$, then the integral over $t$ equals $$-\sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \frac{\log(c)^i}{c^{k-1}x_1\cdots x_{n-1}} + \sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \frac{\log(x_1\cdots x_{n-1})^i}{(x_1\cdots x_{n-1})^k}$$
Hence, if $b\leq\frac{c}{a}$ and $k\neq 1$, we get a recurrence formula: $$I_n^{k,l}(b,c) = -\sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \left( \sum_{j=0}^i \binom{i}{j} \frac{\log(a)^{i-j}}{a^{k-1}} I_{n-1}^{k,j}(1,\frac{c}{a}) - \frac{\log(b)^i}{b^{k-1}} I_{n-1}^{1,0}(1,b) - I_{n-1}^{k,i}(b,\infty) + \frac{\log(c)^i}{c^{k-1}}I_{n-1}^{1,0}(\frac{c}{a},\infty) \right).$$ It can be easily verified that the same formula holds in the case $b\geq\frac{c}{a}$ as well.
Similarly, when $k=1$, we get the formula: $$I_n^{1,l}(b,c) = \frac{1}{l+1}\left(\sum_{j=0}^{l+1} \binom{l+1}{j} \log(a)^{l+1-j} I_{n-1}^{1,j}(1,\frac{c}{a}) - \log(b)^{l+1} I_{n-1}^{1,0}(1,b) - I_{n-1}^{1,l+1}(b,\infty) + \log(c)^{l+1} I_{n-1}^{1,0}(\frac{c}{a},\infty) \right).$$
These formulae allow to compute the integral $I_n^{k,0}(b,c)$ recursively.