Bounds on $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$

This is a twice the bound of the OP:

Suppose $c_n\searrow0$.

Fix $M\in\mathbb{N}$ and for $n\geq N$ let $$P_n(x)=\sum^n_{k=M}e^{ikx}=\frac{e^{iMx}-e^{i(n+1)}}{1-e^{ix}}=e^{iMx}\frac{1-e^{(n-M+1)ix}}{1-e^{ix}}=\frac{\sin\big((n-M+1)x/2\big)}{\sin(x/2)}e^{i(n-M)x/2}$$ From this, we get that $$ S_n(x)=\sum^n_{k=M}\sin kx =\frac{\sin\big((n-M+1)x/2\big)}{\sin(x/2)}\sin\big((n-M)x/2\big) $$

Then by summation by parts $$\begin{align} \sum^N_{n=M} c_n e^{inx} &=c_NP_N(x) +\sum^{N-1}_{n=M}(c_n-c_{n+1})P_n(x)\\ \sum^N_{n=M} c_n \sin nx & = c_NS_N(x) +\sum^{N-1}_{n=M}(c_n-c_{n+1})S_n(x) \end{align}$$ Then $$ \begin{align} \Big|\sum^N_{n=M} c_n e^{inx}\Big| &\leq c_N|P_N(x)| +\sum^{N-1}_{n=M}(c_n-c_{n+1})|P_n(x)|\\ \Big|\sum^N_{n=M} c_n \sin nx\Big| &\leq c_N|S_N(x)| +\sum^{N-1}_{n=M}(c_n-c_{n+1})|S_n(x)| \end{align} $$

For $P_n$ and $S_n$ we have the estimates $$\begin{align} |P_n(x)|&= \big|\sin\big((n-M+1)x/2\big)\big|\csc(x/2)|\leq |\csc(x/2)| \\ |S_n(x)|&=\Big|\sin\big((n-M+1)x/2\big)\sin\big((n-M)x/2\big)\Big||\csc(x/2)|\leq|\csc(x/2)| \end{align}$$ for all $x\neq0$ and $n$,

\begin{align} \Big|\sum^N_{n=M} c_n e^{inx}\Big|&\leq \Big(c_N +\sum^{N-1}_{n=M}(c_n-c_{n+1})\Big)|\csc(x/2)|\leq c_M |\csc(x/2)|\tag{1}\label{one}\\ \Big|\sum^N_{n=M} c_n \sin nx\Big|&\leq \Big(c_N +\sum^{N-1}_{n=M}(c_n-c_{n+1})\Big)|\csc(x/2)|\leq c_M|\csc(x/2)|\tag{2}\label{two} \end{align}

For the OP's series $$ \sum^\infty_{n=1}\frac{\sin nx}{n}=\frac{1}{2i}\sum_{|n|\geq1}\operatorname{sign}(n)\frac{e^{inx}}{n}=\frac12(\pi-x),\qquad 0<x<2\pi $$

From \eqref{one} or \eqref{two} we obtain bound $$\Big|\sum^{N-1}_{n=1}\frac{\sin nx}{n}-\frac{\pi-x}{2}\Big|=\Big|\sum^\infty_{n=N}\frac{\sin nx}{n}\Big|\leq \frac{1}{N}|\csc(x/2)| $$

For $x=1$, we get twice the OP's bound. I ignore at the moment how to improve the bound in \eqref{one} by a factor of $1/2$ which is what the OP is observing.

Note that $\displaystyle\left|\sum_{k=1}^n \frac{\sin k}k - \frac12(\pi-1)\right| = \left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|$.

We prove the following inequality:

$$ \left|\sum_{k=n+1}^\infty \frac{\sin k}k \right| \leq \frac{1}{2 \sin(\frac 12)} \frac 1{n+1} + \frac{3}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)} $$

Let $\displaystyle S_n = \sum_{k=1}^n \sin k = \frac{\cos(1/2)-\cos(n+\frac 12)}{2\sin(\frac 12)}$ and $\displaystyle T_n = \sum_{k=1}^n S_k = \frac{n}{2\tan(\frac 12)} - \frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)}$.

Performing summation by parts twice yields

$$\sum_{k=n+1}^\infty \frac{\sin k}k = \sum_{k=n+1}^\infty \Big(\frac{S_k}{k(k+1)}\Big) - \frac{S_n}{n+1} =\sum_{k=n+1}^\infty \Big(\frac{2T_k}{k(k+1)(k+2)}\Big) - \frac{T_n}{(n+1)(n+2)} - \frac{S_n}{n+1}. $$

Note that $$\sum_{k=n+1}^\infty \frac{2T_k}{k(k+1)(k+2)} = \frac{1}{\tan(\frac 12)} \frac{1}{n+2} - \frac{1}{2 \sin^2(\frac 12)} \sum_{k=n+1}^\infty \frac{\sin(k+1)-\sin(1)}{k(k+1)(k+2)} $$ and

$$\begin{align} \frac{T_n}{(n+1)(n+2)} + \frac{S_n}{n+1} = &\phantom{-}\Big(\frac{1}{\tan(\frac 12)}-\frac{\cos(n+\frac 12)}{2 \sin(\frac 12)} \Big) \frac 1{n+1} \\ &- \Big(\frac{1}{\tan(\frac 12)} + \frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)} \Big) \frac 1{(n+1)(n+2)}. \end{align} $$

Therefore $$\begin{aligned} \sum_{k=n+1}^\infty \frac{\sin k}k = &\phantom{+}\frac{\cos(n+\frac 12)}{2 \sin(\frac 12)} \frac 1{n+1} + \frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)} \\ &-\frac{1}{2 \sin^2(\frac 12)} \sum_{k=n+1}^\infty \frac{\sin(k+1)-\sin(1)}{k(k+1)(k+2)} \end{aligned} $$

and

$$\begin{aligned} \left|\sum_{k=n+1}^\infty \frac{\sin k}k \right| &\leq \frac{|\cos(n+\frac 12)|}{2 \sin(\frac 12)} \frac 1{n+1} + \frac{|\sin(n+1)-\sin (1)|+1}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)} \\ &\leq \frac{1}{2 \sin(\frac 12)} \frac 1{n+1} + \frac{3}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)} \end{aligned} $$

A slightly worse estimate may be obtained as follows. If $|\arg z|<\pi$ then $$ \log (1 + z) = \sum\limits_{k = 1}^N {( - 1)^{k + 1} \frac{{z^k }}{k}} + ( - 1)^N z^{N + 1} \int_0^1 {\frac{{t^N }}{{1 + zt}}dt} $$ for any $N\geq 0$. Substituting $z = - e^i$, taking the imaginary part of each side and rearranging, we find $$ \sum\limits_{k = 1}^N {\frac{{\sin (k)}}{k}} - \frac{{\pi - 1}}{2} = \Im \left[ { - e^{i(N + 1)} \int_0^1 {\frac{{t^N }}{{1 - e^i t}}dt} } \right]. $$ Here \begin{align*} \left| {\Im \left[ { - e^{i(N + 1)} \int_0^1 {\frac{{t^N }}{{1 - e^i t}}dt} } \right]} \right| & \le \int_0^1 {\frac{{t^N }}{{\left| {1 - e^i t} \right|}}dt} \le \frac{1}{{\sin 1}}\int_0^1 {t^N dt} \\ & = \frac{1}{{(N + 1)\sin 1}} < \frac{{1.1884}}{{N + 1}}. \end{align*} This is larger than the expected result by a factor of $\frac{1}{{\cos (1/2)}}=1.13949\ldots$. An exact formula for the remainder can be obtained by computing the imaginary part explicitly, but it results in a complicated formula which does not seem to be useful for better estimates.

Addendum. From the link you posted, the remainder term is $$ {\int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)\sin \left( {\left( {n + \tfrac{1}{2}} \right)t} \right)dt} - \int_{n + 1/2}^\infty {\frac{{\sin t}}{t}dt} }. $$ Integration by parts gives $$ \int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)e^{i\left( {n + \frac{1}{2}} \right)t} dt} = ie^{i\left( {n + \frac{1}{2}} \right)} \left( {1 - \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}} \right)\frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right). $$ Taking imaginary parts $$ \int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)\sin \left( {\left( {n + \tfrac{1}{2}} \right)t} \right)dt} = \left( {1 - \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}} \right)\frac{{\cos \left( {n + \frac{1}{2}} \right)}}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right). $$ By another integration by parts $$ \int_{n + 1/2}^\infty {\frac{{\sin t}}{t}dt} = \frac{1}{n}\cos \left( {n + \tfrac{1}{2}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right). $$ Thus, your remainder term is $$ -\frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}\frac{1}{n}\cos \left( {n + \tfrac{1}{2}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right), $$ which for large $n$ is at most $$ \approx \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}\frac{1}{n} $$ in absolute value. Because of the big-$\mathcal{O}$, in principle, it could be slightly larger, so this is not a complete proof.