Show $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = \|f\|_{\infty}$ for $f \in L^{\infty}$
The result holds as soon as $\|f\|_\infty$ is positive and finite.
To prove this, assume without loss of generality that $f\geqslant0$ almost everywhere and $\|f\|_\infty=1$. Then $0\leqslant f^{n+1}\leqslant f^n$ almost everywhere hence $0\leqslant a_{n+1}\leqslant a_n$. Since $a_{n}\ne0$, this yields $$ \limsup\limits_{n\to\infty}\ a_{n+1}/a_n\leqslant1. $$ In the other direction, note that for every positive $u\lt v\lt1$, $A=[f\geqslant u]$ and $B=[f\geqslant v]$ both have positive measure, and that, for every $n\geqslant0$, $$ a_n\geqslant\int_Bf^n\geqslant v^n\mu(B). $$ Hence, $$ a_{n+1}\geqslant u\int_A f^n=ua_n-u\int_{X\setminus A}f^n\geqslant ua_n-\mu(X\setminus A)u^{n+1}, $$ where the first inequality comes from the fact that $f^{n+1}\geqslant uf^n$ on $A$ and $f^{n+1}\geqslant 0$ everywhere, and the second inequality comes from the fact that $f^n\lt u^n$ on $X\setminus A$.
Together, these two lower bounds on $a_n$ and $a_{n+1}$ yield $$ \frac{a_{n+1}}{a_n}\geqslant u-\frac{u \cdot \mu(X\setminus A)}{\mu(B)}\left(\frac{u}v\right)^n. $$ Since $u\lt v$ and $\mu(X\setminus A)$ is finite, $\liminf\limits_{n\to\infty}\ a_{n+1}/a_n\geqslant u$. This holds for every $u\lt1$, hence $$ \lim\limits_{n\to\infty}\ a_{n+1}/a_n=1. $$
(Note added 2017.04.23 by @Did.) The note below by the OP describes incorrectly the trouble with this answer. To be brief, not much can be saved from this post and the lack of desire of this OP and of the asker (both still present on the site) to correct the situation is flabbergasting. For more details please see the comments thread.
Note added: I make a mistake here. First I thought that $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=L$ would imply $\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}=L$. But this is not correct. Please refer to the proof by @Did.
First it's easy to see that $$(a_n)^{\frac{1}{n}}=\Big(\int_X|f|^nd\mu\Big)^{\frac{1}{n}}\leq\|f\|_\infty\mu(X)^\frac{1}{n},$$ which implies that $$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}\leq\|f\|_\infty,$$ where we have used the fact that $\mu(X)$ is finite. On the other hand, by definition of $\|f\|_\infty$, for all $\epsilon>0$, there exists a measurable set $E$ in $X$ such that $\mu(E)>0$ and $f\geq \|f\|_\infty-\epsilon$ on $E$. Hence, we have $$(a_n)^{\frac{1}{n}}=\Big(\int_X|f|^nd\mu\Big)^{\frac{1}{n}}\geq\Big(\int_E|f|^nd\mu\Big)^{\frac{1}{n}}=(\|f\|_\infty-\epsilon)\mu(E)^{\frac{1}{n}}.$$ As $n\rightarrow\infty$, we have $$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}\geq(\|f\|_\infty-\epsilon).$$ Combining the above inequalities, we have $$\|f\|_\infty\geq\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}\geq(\|f\|_\infty-\epsilon).$$ Since $\epsilon>0$ is arbitrary, we have $$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}=\|f\|_\infty.$$
Now the result follows easily from the fact that $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}}.$$
Disclaimer. This is not an answer, but rather something "too long for a comment" that provides what is missing to make Paul's proof complete.
Paul has shown that $a_n^{1/n}\to \Vert f\Vert_\infty$. As observed above by Jonas, it is enough to show that the sequence $c_n:=\frac{a_{n+1}}{a_n}$ is convergent, because then its limit $L$ will be the same as that of $a_n^{1/n}$.
First note that $a_{n+1}=\int_X \vert f\vert^{n+1}\, d\mu\leq \Vert f\Vert_\infty\,\int_X \vert f\vert^n \, d\mu=\Vert f\Vert_\infty\, a_n$, so that $c_n$ is bounded above by $\Vert f\Vert_\infty$. Hence, it is enough to show that the sequence $(c_n)$ is nondecreasing. In other words, one has to show that $$a_{n+1}^2\leq a_n\, a_{n+2}\, . $$ But this follows from Cauchy-Schwarz's inequality: $$a_{n+1}=\int_X\vert f\vert^{n+1}\, d\mu =\int_X \vert f\vert^{\frac{n}2}\,\vert f\vert^{\frac{n+2}2}\, d\mu\leq \left(\int_X\vert f\vert^n d\mu\right)^{1/2}\left(\int_X\vert f\vert^{n+2} d\mu\right)^{1/2}=a_n^{1/2}a_{n+2}^{1/2}\, . $$