Your conjecture is actually mistaken, unfortunately - due to the visualization you've got, you're missing out on seeing the (quite small) oscillation of the $\sin x$. Although it's bounded, it's not negligible; rather, that actually is the limit function. If you fix $x$, then
$$\lim_{n \to \infty} |f_n(x) - \sin x| = \lim_{n \to \infty} \frac{|x|^3}{n^2} = |x|^3 \lim_{n \to \infty} \frac 1 {n^2} = |x|^3 \cdot 0 = 0$$
for any fixed $x$.
As far as uniform convergence, you can see from the above computation that the choice of $n$ to make the error $|x|^3 / n^2$ less than $\epsilon$ does depend on $x$, and cannot be uniformly bounded for all $x$ - hence, the sequence is not uniformly convergent on $\mathbb{R}$.
It is uniformly convergent on any bounded interval, though. Taking $[-10, 10]$ for one, we have that
$$|f_n(x) - \sin x| \le \frac{1000}{n^2}$$
is a bound independent of $x$. Choosing $N$ on the order $\sqrt{\epsilon / 1000}$ then gives
$$n \ge N \implies |f_n(x) - \sin x| < \epsilon$$
for all $x \in [-10, 10]$, which implies uniform convergence.
Best Answer
In fact since the limit function $f(x)$ is equal to zero every where we have$$d_n=\sup_{\Bbb R} |f_n(x)-f(x)|$$The uniform convergence states that $$\forall \epsilon >0\quad,\quad\exists N\quad,\quad n>N\implies d_n<\epsilon$$which is impossible since $d_n\to {1\over \sqrt e}$ as is proved.
P.S.
$$\lim_{n\to \infty}\left(\sqrt{ \frac{n}{n+1}}\right)^n{=\exp\left(\lim_{n\to \infty}{n\over 2}\ln{n\over n+1}\right)\\=\exp\left(\lim_{u\to 0}{1\over 2u}\ln{1\over u+1}\right)\\=\exp\left(\lim_{u\to 0}-{1\over 2}{1\over u+1}\right)\\={1\over \sqrt e}}$$where we have used the L^Hopital's rule.