Yes indeed. You may assume that the complexes belong to a Grothendieck category if you wish, or just stick with abelian groups. (Grothendieck category: abelian category with exact filtered direct limits that possesses a generator). In the countable case, you have Milnor's exact sequence

$$
\oplus_{n} A_n \longrightarrow \oplus_{n} A_n \longrightarrow \lim_{n} A_n
$$

The first arrow, usually denoted by $1-shift$ takes an element in $A_n$ and sends it to itself in $A_n$ and to the negative of its image by the transition map in $A_{n+1}$. It is an exercise that $1-shift$ is injective on abelian groups (and the proof can be adapted to the general situation).

You have a similar sequence for the $B_n$'s and a map of exact sequences given by $\oplus_{n} f_n$ in the first and second factor and the induced map in the third. When you interpret this diagram in the derived catdegory it is clear that $\oplus_{n} f_n$ is a quasi-isomorphism and your diagram of exact sequences becomes a diagram of triangles. This implies that the induced map is a quasi-isomorphism.

For certain more general systems there are some results in my joint paper with Jeremías & Souto, here or in the CJM site. You can find here an exposition of homotopy colimitis in the context of complexes. Hope this helps.