We are in the category of $R$-modules. Let us consider an inverse system $\{M_n^\bullet\}_{n\geq 1}$ where each $M_n^\bullet$ is an exact sequence.
$$ \dots \longrightarrow M_n^{i-1}\longrightarrow M_n^i \longrightarrow M_n^{i+1}\longrightarrow \dots $$
Suppose that for any given $i\in \mathbb Z$, the maps of the inverse system are $M_{n+1}^i\longrightarrow M_n^i$ are all surjections (in my situation you could even take them to be split surjections).
In other words, for each fixed $i$, the inverse system $\{M_n^i\}_{n\geq 1}$ certainly satisfies the Mittag-Leffler condition.
Now, if we set $M^i:=\varprojlim_{n\geq 1}M_n^i$ for each $i$, is the sequence $$ \dots \longrightarrow M^{i-1}\longrightarrow M^i \longrightarrow M^{i+1}\longrightarrow \dots $$ still exact? I know this is true if the sequences were just short exact sequences, but I am looking for an answer for arbitrary unbounded complexes.
This follows from the case of short exact sequences.
Let $B^i_n$ be the image of $M^{i-1}_n\to M^i_n$. Then the map $B^i_n\to B^i_{n-1}$ induced by $M^i_n\to M^i_{n-1}$ is surjective, and so, by Mittag-Leffler, the inverse limit of the short exact sequences $$0\to B^i_n\to M^i_n\to B^{i+1}_n\to0$$ is exact.
But the inverse limit of the sequences $M^\bullet_n$ is obtained by splicing the inverse limits of these short exact sequences, and so that is also exact.