The answer is no. E.g., let $n=3$ and $z_j=e^{i(j-1)2\pi/3}$ for $j=1,2,3$. Then $f(z)>|z|+15/100>|z|$ if $|z|\le1$.
This counterexample generalizes to any $n\ge3$. Indeed, take any $n\ge3$ and let $z_j=e^{i(j-1)2\pi/n}$ for $j=1,\dots,n$. Then
$$f(z)=\frac1n\,\sum_{j=1}^n f_j(z),$$
where $f_j(z):=|z-z_j|$.
Note that for each $j$
the function $f_j$ is convex and, moreover, $f_j$ is strictly convex on any straight line not through the point $z_j$. Since $n\ge3$, there is no straight line passing through all the points $z_1,\dots,z_n$. So, the function $f=\frac1n\,\sum_{j=1}^n f_j$ is strictly convex.
Also, for any real $a$ and $b$, the derivative of $f$ at $0$ along the vector $(a,b)=a+ib$ is
$$\begin{aligned}&\frac d{ds}\,f(s(a+ib))\Big|_{s=0} \\
&=-\frac1n\,(a,b)\cdot\sum_{j=1}^n
\Big(\cos\frac{(j-1)2\pi}n,\sin\frac{(j-1)2\pi}n\Big) \\
&=-\frac1n\,(a,b)\cdot(0,0)=0,
\end{aligned}$$
where $\cdot$ denotes the dot product.
Thus, the strictly convex function $f$ has a unique minimum at $0$, and the minimum value of $f$ is $1$. That is, $f(0)=1<f(z)$ for all $z$ such that $0<|z|\le1$.
So, for all $z$ such that $0<|z|\le1$, we have
$f(z)>1\ge|z|$ and hence $f(z)\ne|z|$. Also, $f(0)=1\ne0$. We conclude that there is no $z$ such that $|z|\le1$ and $f(z)=|z|$, which proves the claim.
On the other hand, the answer is yes for $n=1$ and $n=2$.
