The limit doesn't always exist.
For convenience I will use the Lebesgue measure; similar construction can also be done for most $\mu$.
Let $c_n$ be a sequence of increasing positive integers. Consider the function $$ f(x) = \sum_{n = 0}^\infty c_n^2 \chi_{[n, n + (c_n!)^{-1}]}(x) $$
By the disjoint support we have that $$ \|f\|_{L^p}^p = \sum_{n = 0}^\infty \frac{(c_n)^{2p}}{c_n !} $$
Let us choose $c_n$ so that $$ c_{n+1} \geq 10 (c_n)^{2n} $$
Lim-sup
If $p = c_N$ for some $N$, we have $$ \|f\|_{L^p}^p \geq \frac{(c_N)^{2c_N}}{c_N!} \geq p^p $$ This implies $$ \limsup_{p \to \infty} \frac{\ln \|f\|_p}{\ln p} \geq 1$$
Lim-inf
Let $p = (c_N)^{2N}$ for some $n$. We have $$ \sum_{n = 0}^N \frac{(c_n)^{2p}}{c_n!} \leq (c_N)^{2p} \cdot e = e\cdot p^{p/N}$$ On the other hand $$ \sum_{n = N+1}^\infty \frac{(c_n)^{2p}}{c_n!} \leq \sum_{n = N+1}^\infty \frac{(c_n)^{2p}}{c_n (c_n - 1) \cdots (c_n - 2p + 1) \cdot (c_n - 2p)!} $$ Noting that $2p \leq \frac15 c_{N+1}$ we have $$ \leq \sum_{n = N+1}^\infty \left( \frac54\right)^{2p} \cdot \frac{1}{(c_n - 2p)!} \leq \left( \frac54 \right)^{2p} \cdot e $$ So $$ \|f\|_{L^p} \leq e^{1/p} \left( p^{1/N} + \frac{25}{16}\right) $$ For all sufficiently large $N$, using that $p^{1/N} = (c_N)^2 > 2$ we have $$ \ln \|f\|_p \leq \frac{1}{p} + \ln 2 + \frac{1}{N} \ln p $$ and hence $$ \liminf_{p\to \infty} \frac{\ln \|f\|_p}{\ln p} = 0 $$
Remarks
Note that the Lebesgue measure of the support of $f$ is has size less than $e$. So the same construction would work even if your measure is finite. What it really needs is enough sets of arbitrarily small measure.
To illustrate this final fact, let us consider the special case of the counting measure on $\mathbb{N}$; in other words, let's look at the $\ell_p$ norms on real sequences. (The argument here is a modification of one given by Alexandre Eremenko.)
By Minkowski's inequality we know that that $\ell_p$ norms are decreasing in $p$. So if we consider the mapping $\psi: [0,1]\ni x \mapsto \ln \|f\|_{\ell_{1/x}}$ for any fixed sequence $f$, we see that $\psi$ is
- An increasing function of $x$ (by Minkowski)
- A convex function of $x$ (by Riesz-convexity)
Now the function $y\mapsto e^{-y}$ is convex, and hence the function $\phi: y\mapsto \psi(e^{-y})$ is also convex, as it is the composition of two convex functions, the outer of which is increasing.
Note that $\phi(\ln p) = \ln \|f\|_{\ell_p}$.
We can conclude that the desired limit exists in this case since for any convex function $\phi$, the limit $\lim_{x\to\infty} \phi(x) / x$ exists in the sense given in the question (basically since $\phi'$ is increasing).