Assume that $\int_{\Omega}-\log|f|d\mu<\infty$. Let $g(p):=\frac 1p\log\int_{\Omega}|f|^pd\mu-\int_{\Omega}\log|f|d\mu$.
Since $t\mapsto \log t$ is concave, by Jensen inequality we get $g(p)\geqslant 0$. Using the inequality $\ln(1+t)\leqslant t$ we have
$$0\leqslant g(p)\leqslant \frac 1p\left(\int_{\Omega}|f|^pd\mu-1\right)-\int_{\Omega}\log|f|d\mu.$$
Now the problem reduces to show that $\lim_{p\to 0}\frac 1p\left(\int_{\Omega}|f|^pd\mu-1\right)-\int_{\Omega}\log|f|d\mu=0$. To see that, take a sequence $\{p_n\}$ which converges to $0$ and put $f_n(x):=\frac{|f(x)|^{p_n}-1}{p_n}-\log |f(x)|$.
The sequence $\{f_n\}$ converges almost everywhere to $0$ and we have, if $t\geq 1$, $0
Now assume that $\int_{\Omega}\log|f|d\mu=-\infty$. Consider $f_R:=|f|\mathbf 1_{\{|f|\gt 1/R\}}$. Then $-\log |f_R|\leqslant \log R$, hence by the previous case,
$$\tag{*} \lim_{p \to 0}\left[ \int_{\Omega}\left|f_R\right|^pd\mu \right]^{\frac{1}{p}}=\exp\left(\int_\Omega\log|f_R|\mathrm \mu\right).$$
Fix a positive $\varepsilon$ and by monotone convergence, we may choose $R_0$ such that $\exp\left(\int_\Omega\log|f_{R_0} |\mathrm \mu\right)\lt \varepsilon$ and $1/R_0\lt \varepsilon$. Then
$$\left(\int_{\Omega}|f|^p\mathrm d\mu\right)^{1/p} \leqslant
\frac 1{R_0}+ \left[ \int_{\Omega}\left|f_{R_0} \right|^pd\mu \right]^{\frac{1}{p}},$$
so that
$$\limsup_{p\to 0}\left(\int_{\Omega}|f|^p\mathrm d\mu\right)^{1/p} \leqslant
\frac 1{R_0}+\exp\left(\int_\Omega\log|f_{R_0} |\mathrm \mu\right)\leqslant 2\varepsilon.$$