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 $n\to\infty$.

In general, the answer is no. Moreover, the answer is no even if \begin{equation} \phi(t)=t\ln(1+t). \tag{1} \end{equation}

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{2\phi(b)}, \end{equation*} $\phi$ is as given by (1), 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 $E\phi(|Z_i|)=1$, so that $\|Z_i\|_\phi\le1$. On the other hand, for all real $c>0$ and all natural $n\ge2$ \begin{equation*} \begin{aligned} &E\phi\Big(\Big|\frac1n\sum_{i=1}^n Z_i\Big|/c\Big) \\ &\ge\sum_{i=1}^n \phi\Big(\frac b{cn}\Big)P(|Z_i|=b,\ Z_j=0\ \forall j\ne i) \\ &=n \phi\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 $E\phi\big(\big|\frac1n\sum_{i=1}^n Z_i\big|/c\big)>1$ and hence $\|\frac1n\sum_{i=1}^n Z_i\|_\phi\ge c$ and hence \begin{equation*} \Big\|\frac1n\sum_{i=1}^n Z_i\Big\|_\phi\not\to0 \end{equation*} as $n\to\infty$.

More generally, the answer will remain no if $\phi(t)=t \ell(t)$, where $\ell$ is any function such that $\ell(t)$ is slowly varying as $t\to\infty$. Yet more generally, the answer will remain no if $\phi(t)=t L(t)$, where $L$ is any function such that $\sup\limits_{K\in(0,\infty)}\limsup\limits_{t\to\infty}\dfrac{L(Kt)}{L(t)}<\infty$.

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} \|\frac1n\sum_{i=1}^n Z_i\|_f\not\to0 \end{equation} as $n\to\infty$.

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 $n\to\infty$.