**UPDATE.** Formulae are generalized to include the case $k=1$.

Let us assume that $a$ is fixed, and define:
$$
I_n^{k,l}(b,c) := \iiiint_{[1,a]^n} \frac {\chi(\{ b \le x_1 \cdots x_n \le c \})\log(x_1 x_2 \cdots x_n)^l} {( x_1 x_2 \cdots x_n)^k} \,dx_1 \cdots dx_n.
$$
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 following formula, which holds for all $l\geq 0$:
$$\int \frac{\log(y)^l}{y^k}dy = \begin{cases}
\frac{\log(y)^{l+1}}{l+1}, & \text{if }k=1;\\
-\sum_{i=0}^l \frac{l!}{i!(k-1)^{l+1-i}} \frac{\log(y)^i}{y^{k-1}} & \text{if }k\ne 1.
\end{cases}$$


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.