As Robin Chapman mentions in his comment, the difficulty of investigating
the convergence of
$$
\sum_{n=1}^\infty\frac1{n^3\sin^2n}
$$
is due to lack of knowledge about the behavior of $|n\sin n|$ as $n\to\infty$,
while the latter is related to rational approximations to $\pi$ as follows.
Neglecting the terms of the sum for which $n|\sin n|\ge n^\varepsilon$
($\varepsilon>0$ is arbitrary),
as they all contribute only to the `convergent part' of the sum, the question
is equivalent to the one for the series
$$
\sum_{n:n|\sin n|< n^\varepsilon}\frac1{n^3\sin^2n}.
\qquad(1)
$$
For any such $n$, let $q=q(n)$ minimizes the distance $|\pi q-n|\le\pi/2$.
Then
$$
\sin|\pi q-n|=|\sin n|< \frac1{n^{1-\varepsilon}},
$$
so that $|\pi q-n|\le C_1/n^{1-\varepsilon}$ for some absolute constant $C_1$
(here we use that $\sin x\sim x$ as $x\to0$). Therefore,
$$
\biggl|\pi-\frac nq\biggr|<\frac{C_1}{qn^{-\varepsilon}},
$$
equivalently
$$
\biggl|\pi-\frac nq\biggr|<\frac{C_2}{n^{2-\varepsilon}}
\quad\text{or}\quad
\biggl|\pi-\frac nq\biggr|<\frac{C_2'}{q^{2-\varepsilon}}
$$
(because $n/q\approx\pi$) for all $n$ participating in the sum (1).
It is now clear that the convergence of the sum (1) depends
on how often we have
$$
\biggl|\pi-\frac nq\biggr|<\frac{C_2'}{q^{2-\varepsilon}}
$$
and how small is the quantity in these cases. (Note that
it follows from Dirichlet's theorem that an even stronger inequality,
$$
\biggl|\pi-\frac nq\biggr|<\frac1{q^2},
$$
happens for infinitely many pairs $n$ and $q$.)
The series (1) converges if and only if
$$
\sum_{n:|\pi-n/q|< C_2n^{-2+\varepsilon}}\frac1{n^5|\pi-n/q|^2}
$$
converges. We can replace the summation by summing over $q$
(again, for each term $\pi q\approx n$) and then sum the result
over all $q$, because the
terms corresponding to $|\pi-n/q|< C_2n^{-2+\varepsilon}$ do not
influence on the convergence:
$$
\sum_{q=1}^\infty\frac1{q^5|\pi-n/q|^2}
=\sum_{q=1}^\infty\frac1{q^3(\pi q-n)^2}
\qquad(2)
$$
where $n=n(q)$ is now chosen to minimize $|\pi-n/q|$.
Summarizing, the original series converges if and only if the series in (2)
converges.
It is already an interesting question of what can be said about the
convergence of (2) if we replace $\pi$ by other constant $\alpha$,
for example by a "generic irrationality". The series
$$
\sum_{q=1}^\infty\frac1{q^3(\alpha q-n)^2}
$$
for a real quadratic irrationality $\alpha$ converges because the best
approximations are $C_3/q^2\le|\alpha-n/q|\le C_4/q^2$, and they are achieved on
the convergents $n/q$ with $q$ increasing geometrically. A more delicate
question seems to be for $\alpha=e$, because one third of its convergents satisfies
$$
C_3\frac{\log\log q}{q^2\log q}<\biggl|e-\frac pq\biggr|< C_4\frac{\log\log q}{q^2\log q}
$$
(see, e.g., [C.S.Davis, Bull. Austral. Math. Soc. 20 (1979) 407--410]).
The number $e$, quadratic irrationalities, and even algebraic numbers
are `generic' in the sense that their irrationality exponent is known to be 2.
What about $\pi$?
The irrationality exponent $\mu=\mu(\alpha)$ of a real irrational number
$\alpha$ is defined as the infimum of exponents $\gamma$
such that the inequality $|\alpha-n/q|\le|q|^{-\gamma}$ has
only finitely many solutions in $(n,q)\in\Bbb Z^2$ with $q\ne0$.
(So, Dirichlet's theorem implies that $\mu(\alpha)\ge2$. At the same
time from metric number theory we know that it is 2 for almost all real irrationals.)
Assume that $\mu(\pi)>5/2$, then there are infinitely many
solutions to the inequality
$$
\biggl|\pi-\frac nq\biggr|<\frac{C_5}{q^{5/2}},
$$
hence infinitely many terms in (2) are bounded below by $1/C_5$, so that
the series diverges (and (1) does as well). Although the general
belief is that $\mu(\pi)=2$, the best known result of V.Salikhov (see
this answer
by Gerry and my comment)
only asserts that $\mu(\pi)<7.6064\dots$,.
I hope that this explains the problem of determining the behavior of the series in question.
A rather detailed discussion of the subject can be found in Knopp's Theory and Application of Infinite Series (see § 41, pp. 298-305). He mentiones that the idea of a possible boundary between convergent and divergent series was suggested by du Bois-Reymond. There are many negative (and mostly elementary) results showing that no such boundary, in whatever sense it might be defined, can exist.
Stieltjes observed that for an arbitary monotone decreasing sequence $(\epsilon_n)$ with the limit $0$, there exist a convergent series $\sum c_n$ and a divergent series $\sum d_n$
such that $c_n=\epsilon_nd_n$. (This can be easily deduced from the Abel-Dini theorem).
Pringsheim remarked that, for a convergent and a divergent series with positive terms, the ratio $c_n/d_n$ can assume all possible values, since one may have simultaneously
$$\liminf\frac{c_n}{d_n}=0\qquad\mbox {and}\qquad\limsup\frac{c_n}{d_n}=\infty.$$
I like the following geometric interpretation. Given a (convergent or divergent) series $\sum a_n$, let's mark the sequence of points $(n,a_n)\in\mathbb R^2$
and join the consecutive points by straight segments. Then there is a convergent series $\sum c_n$ and a divergent series $\sum d_n$ (both with positive and monotonically decreasing terms) such that the corresponding polygonal graphs can intersect in an indefinite number of points.
The results remain essentially unaltered even if one requires that both sequences $(c_n)$ and $(d_n)$ are fully monotone, which is a very strong monotonicity assumption. This was shown by Hahn ("Über Reihen mit monoton abnehmenden Gliedern", Monatsh. für Math., Vol. 33 (1923), pp. 121-134).
Best Answer
I think the answer is no.
By functional analysis, we know that it suffices to show the existence of a sequence $a_n$ such that $\lVert a \rVert_2 = 1$ and $$\sum_{n = 1}^{\infty} \left(\sum_{j = 1}^\infty \frac{a_{jn}}{j}\right)^2$$ is arbitrarily large.
Take the first $k$ prime numbers $p_1, \cdots, p_k$, and an integer $s\ge1$. Let $\mathbb{B}_{k, s}$ denote the set $$\mathbb{B}_{k, s} = \{p_1^{i_1} p_2^{i_2} \cdots p_k^{i_k}: 0 \leq i_1, i_2, \cdots, i_k \leq s\}.$$ We consider the sequence $$a_n = \begin{cases} \frac{1}{\sqrt{|\mathbb{B}_{k, s}|}}\text{ if } n \in \mathbb{B}_{k, s}, \\ 0 \text{ otherwise}. \end{cases}.$$ Then $\lVert{a\rVert}_2 = 1$. However, if $n \in \mathbb{B}_{k, s - 1}$, then $np_i \in \mathbb{B}_{k, s}$ for each $i \in [k]$. So we have $$\sum_{j = 1}^\infty \frac{a_{jn}}{j} \geq \left(\sum_{i = 1}^k \frac{1}{p_i}\right) \cdot \frac{1}{\sqrt{|\mathbb{B}_{k, s}|}}.$$ Thus $$\sum_{n = 1}^{\infty} \left(\sum_{j = 1}^\infty \frac{a_{jn}}{j}\right)^2 \geq \left(\sum_{i = 1}^k \frac{1}{p_i}\right)^2 \frac{|\mathbb{B}_{k, s - 1}|}{|\mathbb{B}_{k, s}|} = \left(\sum_{i = 1}^k \frac{1}{p_i}\right)^2 \cdot \frac{s^k}{(s+1)^k} \gg (\log\log k)^2 \cdot \frac{s^k}{(s+1)^k}.$$ which can be arbitrarily large by tuning $s$ and $k$, as desired.