No. For each $n$ let $(A_{n,m})$ be a sequence of subsets of $\Omega$ each with measure less than $1/n$, but with $\bigcup_m A_{n,m} = \Omega$ (certainly you can do this if $\Omega=[0,1]$ with Lebesgue measure).

Now set  $f^{(m)}_n = \chi_{A_{n,m}}$. Then \[ \lim_n \ \sup_m \|f^{(m)}_n\|_1 = \lim_n \ \sup_m |A_{n,m}| < \lim_n \frac{1}{n} = 0. \] However, for any sequence $(k_n)$ and any $x\in\Omega$, \[ \sup_m |f^{(m)}_{k_n}(x)| = \sup_m \chi_{A_{k_n,m}}(x) = \chi_{\bigcup_m A_{k_n,m}}(x) = 1. \] Thus your conclusion cannot hold.

No. For each $n$ let $(A_{n,m})$ be a sequence of subsets of $\Omega$ each with measure less than $1/n$, but with $\bigcup_m A_{n,m} = \Omega$ (certainly you can do this if $\Omega=[0,1]$ with Lebesgue measure).

Now set 

$f^{(m)}_n = \chi_{A_{n,m}}$. Then

$$ \lim_n \ \sup_m \|f^{(m)}_n\|_1 = \lim_n \ \sup_m |A_{n,m}| < \lim_n \frac{1}{n} = 0. $$

However, for any sequence $(k_n)$ and any $x\in\Omega$,$ \sup_m |f^{(m)}_{k_n}(x)| = \sup_m \chi_{A_{k_n,m}}(x) = \chi_{\bigcup_m A_{k_n,m}}(x) = 1.$

Thus your conclusion cannot hold.