zap, the unit ball of $L_\infty(\mu)$ is not weakly compact (hence not weakly sequentially compact either) because $L_\infty(\mu)$ is not reflexive apart from the trivial cases.
The result you are interested in follows from two facts.
Theorem (Amir–Lindenstrauss). Let $X$ be a weakly compactly generated Banach space. Then weak*-compact subsets of $X^*$ are weak*-sequentially compact.
D. Amir and J. Lindenstrauss, The structure of weakly compact
sets in Banach spaces, Ann. Math. 88 (1968), 35–46.
Fact. Let $\mu$ be a finite measure. Then $L_1(\mu)$ is weakly compactly generated. Indeed, the inclusion map from the reflexive space $L_2(\mu)$ to $L_1(\mu)$ has dense range (and is injective of course).
Observation. We can extend our conclusion to $\sigma$-finite measures $\mu$. Indeed, the Radon–Nikodym theorem yields an isometry from $L_1(\mu)$ to some $L_1(\nu)$ where $\nu$ is a finite measure.