We have $f^n(x,y,z)=(x,\,y+nx,\,z+ny+\tbinom n 2 x)$ Taking $\|\cdot\|_\infty$ as a metric on $(\mathbb R/\mathbb Z)^3$, this implies $d_n(a,b) \le (1+n+n(n-1)/2) \|a-b\|_\infty$ where $d_n$ is the maximum distance between the two orbits $(a,f(a),\dots,f^n(a))$.
So that an $(n,\varepsilon)$-separated set must be $(0,\Omega(\varepsilon/n^2))$-separated (in other words, the metric $d_n$ grows at most quadratically) and therefore, since we are in dimension $d=3$, for fixed $\varepsilon$ its cardinality grows as $O(n^{2d})$, which suffices to conclude that $h_{top}(f)=0$.
Note that the cardinality itself does grow faster than quadratically, as can be seen with the following $n^2(n-1)/2$ points $M_{uv}$: $\left\{\begin{aligned} x=&u/\tbinom n 2\\ y=&v/n\\ z=&0 \end{aligned}\right.$ If $d_n(M_{uv},M_{u'v'})<\varepsilon<1/4$ for some odd $n$, then we have $|n(y'-y)+\tbinom n 2 (x'-x)|<\varepsilon\\ |v'-v+u'-u|<1\\ v'-v = -(u'-u)$ $|(n+1)/2\cdot(y'-y)+\tbinom{(n+1)/2}{2}(x'-x)|<\varepsilon\\ |u'-u|<\frac{n}{(n+1)/4}\varepsilon<1\\ u=u'\\ v=v'$ So that we have a set of $\Omega(n^3)$ points that is $(n,\varepsilon)$-separated.