**Edit:** I improved the constant to $c = \frac{2}{3}$.

**Answer:** Yes, we have
$$
  \inf_{(n_k)} \sup_{i,j \in \mathbb{N}} \|f_{n_i} - f_{n_j}\|_{L^1} \le \frac{2}{3} M, \label{1}\tag{$\ast$}
$$
for each sequence $(f_n)$ in $(L^1)_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{2}{3}$.

To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.

**Lemma.** Three functions $f_1, f_2, f_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances that are all strictly larger than $\frac{2}{3}$.

*Proof.* 
Set $g_1 = |f_1 - f_2|$, $g_2 = |f_1 - f_3|$ and $g_3 = |f_2 - f_3|$.
For any three numbers $r_1,r_2,r_3 \in [0,1]$, the sum of their three mutual distances in $\mathbb{R}$ is at most $2$.

Hence, $\int g_1 + \int g_2 + \int g_3 \le 2$, which shows that it can't happen that all three functions $g_k$ have norm strictly larger than $\frac{2}{3}$. $\square$

*Proof of the claim.*
We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{2}{3}$. 

Then there exists $n_0$ such that $\|f_{n_0} - f_n\|_{L^1} > \frac{2}{3}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f_{n_k})$ such that the supremum in \eqref{1} is no more than $\frac{2}{3}$); let's denote the set of these $n$ by $J$.

For any two $j,k \in J$, it follows from the lemma that $\|f_j - f_k\| \le \frac{2}{3}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f_{n_k})$. Contradiction, since we assumed no such subsequence exists. $\square$

**Remark.** It's easy to see that the constant $\frac{2}{3}$ is optimal for the lemma (divide $[0,1]$ into three distjoint intervals $I_k$ of measure $\frac{1}{3}$ and define $f_k = \mathbf{1} - \mathbf{1}_{I_k}$), but I don't know whether it is optimal for the answer to the question.