Let us show that $f(x)$ is finite.
For each $n\in\mathbb N$ there are at most $nx+1$ fractions $q\in[0,x]$ with denominator $n$, hence their contibution to the sum is $\le \frac{nx+1}{n^3}=\frac x{n^2}+\frac1{n^3}$ so that convergence follows by comparision with the convergent series $\sum\frac 1{n^2}$.
More generally, we conclude for $u<v$ not only that
$$\tag1 f(v)-f(u)\le (v-u)\sum_{n=1}^\infty \frac1{n^2}+\sum_{n=1}^\infty\frac1{n^3}$$
but if we additionally know that all fractions in $[u,v]$ have denominators $\ge N$ we get the stronger estimate
$$\tag2 f(v)-f(u)\le (v-u)\sum_{n=N}^\infty \frac1{n^2}+\sum_{n=N}^\infty\frac1{n^3}.$$
The function $f$ is discontinuous at rational numbers, because there it "jumps" by $\frac1{p(x)^3}$.
Let $\alpha$ be irrational. Let $\epsilon>0$.
Pick $N\in \mathbb N$ such that $\sum_{n=N}^\infty \frac1{n^2}+\sum_{n=N}^\infty\frac1{n^3}<\epsilon$.
Let $u=\max\{\frac {\lfloor k\alpha\rfloor}k\,\mid 1\le k\le N\,\}$, $v=\min\{\frac {\lceil k\alpha\rceil}k\,\mid 1\le k\le N\,\}$.
Then $u<\alpha<v$ and all rationals in $[u,v]$ have denomninator $\ge N$. As $v-u\le 1$, we see from $(2)$ that $f(v)-f(u)<\epsilon$.
Since $f$ is incrreasing, we fave $|f(x)-f(\alpha)|<\epsilon$ for all $x\in(u,v)$, i.e. $f$ is continuous at $\alpha$.