In general, the answer is no. Moreover, the answer is no even if 
$$\phi(t)=f(t):=t\ln(1+t).$$ 

Indeed, suppose that $P(Z_i=0)=1-2p$ and $P(Z_i=b)=p=P(Z_i=-b)$ for all $i$, where 
\begin{equation}
	p:=\frac1{2b\ln(1+b)}
\end{equation}
and $b$ is a large enough positive real number so that $p\in(0,1/2)$. 

Then for all $i$ we have $EZ_i=0$ and $Ef(|Z_i|)=1$, so that $\|Z_i\|_f\le1$. On the other hand, for all real $c>0$ and all natural $n\ge2$
\begin{equation}
\begin{aligned}
	&Ef\Big(\Big|\frac1n\sum_{i=1}^n Z_i\Big|/c\Big) \\ 
	&\ge\sum_{i=1}^n f\Big(\frac b{cn}\Big)P(|Z_i|=b,\ Z_j=0\ \forall j\ne i) \\ 
	&=n f\Big(\frac b{cn}\Big)2p(1-2p)^{n-1} \\ 
	&=\frac{2pb}c\,\ln\Big(1+\frac b{cn}\Big)(1-2p)^{n-1}\to\frac1{2c}>1 
\end{aligned}
\end{equation}
as $n\to\infty$, if $b=n^2$ and $c\in(0,1/2)$. So, for all large enough $n$ we have $Ef\big(\big|\frac1n\sum_{i=1}^n Z_i\big|/c\big)>1$ and hence $\|\frac1n\sum_{i=1}^n Z_i\|_f\ge c$ and hence 
\begin{equation}
	\Big\|\frac1n\sum_{i=1}^n Z_i\Big\|_f\not\to0 
\end{equation}
as $n\to\infty$.