Roughly, the deciding factor in uniform convergence is whether the "rate" of convergence of $f_n(x)$ to its limit is the same for all $x$.
A counterexample: let $f_n(x)=x/n$. The functions $f_n$ converge to $0$ pointwise. However, the time you have to wait for $f_n(x)$ to get within $0.01$ of $0$ depends on $x$. For example, when $x=3$, then the smallest $n$ such that $|f_n(x)|<0.01$ is $n=300$, while for $x=30$, the smallest such $n$ is $n=3000$. As $x$ gets further from $0$, the required $n$ to get $f_n(x)$ within $0.01$ of $0$ gets arbitrarily large, which means $f_n$ does not converge uniformly.
Back to your problem. The function $f_n(x)=\sin(nx)/n^3$ is more complicated, so given $\epsilon$, we cannot find the exact smallest $n$ for which $|f_n(x)|<\epsilon$. However, all we need to do is note that $|\sin nx|\le 1$ always, so that $|f_n(x)|\le 1/n^3$. Note that this bound does not depend on $x$; it gives us a guarantee that after $n$ is large enough, $f_n(x)$ will be close to $0$ for all $x$. Therefore, $f_n$ does indeed converge uniformly.