It is known that when $\mu$ is $\sigma$finte measure, then $L^\infty(\mu)$ is $1$injective. But I want to know whether it is right for any $L^\infty$ spaces.
A useful sufficient condition is that $(X,\Sigma,\mu)$ is a localizable measure space. See 363R of Fremlin, Volume 3, p.308. 


For a compact Hausdorff space $K$ the algebra $C(K)$ is $1$injective iff $K$ is extremally disconnected iff its Boolean algebra of idempotents is complete. So to construct a counterexample we need an $L^\infty$ algebra with an incomplete Boolean algebra of measurable sets modulo null sets. (quite obviously, this Boolean algebra is always $\sigma$complete, and it is also wellknown to be complete if the measure is $\sigma$finite). Now let $X$ be an uncountable set, equipped with the $\sigma$algebra $\mathcal{F}$ that consists of sets that are either countable or have countable complement, and the counting measure $\mu$. Then obviously the only null set is the empty set. Now it's easy to see that the Boolean algebra of countableorcocountable sets is incomplete: if $Y \subset X$ is a subset that is uncountable and has uncountable complement, then the family of all countable subsets of $Y$ has no supremum within $\mathcal{F}$. 

