Here is a partial answer: If $n \geq 2$ is an integer, then
$$ I_n = \frac{1}{(n-1)!2^{n-1}} \sum_{l=0}^{\lfloor n/2 \rfloor} (-1)^l \binom{n}{l} (n-2l)^{n-1} J_{n-2l}, \tag{1} $$
where $J_p$ is defined by
$$ J_{p} = \begin{cases}
\displaystyle \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{\log(2\pi j)}{4j^2-p^2}, & \text{if $p$ is odd}, \\
\displaystyle \frac{\pi}{2}, & \text{if $p$ is even}.
\end{cases} $$
Proof of $\text{(1)}$. The case of even $n$ has already been discussed in other postings, so we focus on odd $n$. We first note that, if $n \geq 1$ is an odd integer, then
\begin{align*}
\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} \sin^n x
&= \frac{1}{(2i)^n} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} \frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} (e^{(n-2l)ix} - e^{-(n-2l)ix}) \\
&= \frac{1}{2^{n-1}} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} (n-2l)^{n-1} \sin((n-2l)x).
\end{align*}
So by applying integration by parts $(n-1)$-times, we get
\begin{align*}
I_n
&= \sum_{k=0}^{\infty} \int_{0}^{\pi} \frac{\sin^n x}{(x+k\pi)^n} \, \mathrm{d}x \\
&= \frac{1}{(n-1)!} \sum_{k=0}^{\infty} \int_{0}^{\pi} \biggl( \frac{1}{x+k\pi} - \frac{1}{(k+1)\pi} \biggr) \biggl( \frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}} \sin^n x \biggr) \, \mathrm{d}x \\
&= \frac{1}{(n-1)!2^{n-1}} \sum_{l=0}^{\frac{n-1}{2}} (-1)^l \binom{n}{l} (n-2l)^{n-1} J_{n-2l},
\end{align*}
where $J_{p}$ is defined by
\begin{align*}
J_{p}
&= \sum_{k=0}^{\infty} \int_{0}^{\pi} \biggl( \frac{1}{x+k\pi} - \frac{1}{(k+1)\pi} \biggr) \sin(px) \, \mathrm{d}x.
\end{align*}
If $p$ is odd, then the above definition is recast as
\begin{align*}
J_{p}
&= \sum_{k=0}^{\infty} \biggl( \int_{0}^{\pi} \frac{1}{x+k\pi} \sin(px) \, \mathrm{d}x - \frac{2}{p\pi(k+1)} \biggr) \\
&= \lim_{N \to \infty} \biggl( \int_{0}^{N \pi} \frac{\sin(p(x \text{ mod } \pi))}{x} \, \mathrm{d}x - \frac{2}{p\pi} H_N \biggr),
\end{align*}
where $H_N = 1 + \frac{1}{2} + \dots + \frac{1}{N}$ is the $N$-th harmonic number. Still assuming that $p$ is an odd integer, Fourier series computation shows that
\begin{align*}
\sin(p(x \text{ mod } \pi))
&= \frac{2}{p\pi} - \frac{4p}{\pi} \sum_{n=1}^{\infty} \frac{\cos(2\pi n x)}{4n^2-p^2} \\
&= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1 - \cos(2\pi j x)}{4j^2-p^2},
\end{align*}
and so,
\begin{align*}
\int_{0}^{N \pi} \frac{\sin(p(x \mathrm{ mod } \pi))}{x} \, \mathrm{d}x
&= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} \int_{0}^{N \pi} \frac{1 - \cos(2 j x)}{x} \, \mathrm{d}x \\
&= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} (\gamma + \log(2\pi j N) - \operatorname{Ci}(2\pi j N) ).
\end{align*}
Plugging this back and using the identity $\frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} = \frac{2}{p\pi}$, which itself follows from the Fourier series of $\sin(p(x \text{ mod } \pi))$, we finally obtain
\begin{align*}
J_{p}
&= \frac{4p}{\pi} \lim_{N \to \infty} \sum_{j=1}^{\infty} \frac{1}{4j^2-p^2} (\gamma + \log(2\pi j N) - \operatorname{Ci}(2\pi j N) - H_N ) \biggr) \\
&= \frac{4p}{\pi} \sum_{j=1}^{\infty} \frac{\log(2\pi j)}{4j^2-p^2}
\end{align*}
as desired. $\square$
Addendum. Here is a Mathematica code for numerical verification of $\text{(1)}$:
n = 5; (* Choose your favorite odd integer >= 3*)
NIntegrate[Evaluate[Sum[1/(x + k Pi)^n, {k, 0, Infinity}] Sin[x]^n], {x, 0, Pi}, WorkingPrecision -> 20]
TermJ[p_] := (4 p)/Pi NSum[Log[2 Pi j]/(4 j^2 - p^2), {j, 1, Infinity}, WorkingPrecision -> 20];
1/((n - 1)! 2^(n - 1)) Sum[Binomial[n, l] (-1)^l (n - 2 l)^(n - 1) TermJ[n - 2 l], {l, 0, (n - 1)/2}]
Clear[n, TermJ];
Best Answer
Using the results listed here we can show that $I_n=2\pi i^n{\cal I}_n(1)$ where ${\cal I}_n$ is the modified Bessel function of the first kind.
Here we will perform the derivation from scratch. Applying the unit circle transformation we have \begin{align}I_n&=\Re\int_0^{2\pi}e^{\sin x}e^{inx}\,dx=\Re\oint_{|z|=1}z^n\exp\left(\frac{z-1/z}{2i}\right)\,\frac{dz}{iz}\end{align} which we can expand as \begin{align}I_n&=\Re\oint_{|z|=1}\sum_{r=0}^\infty\left(\frac i2\right)^r\frac{z^{n-r-1}(1-z^2)^r}{i\cdot r!}\,dz=\Re\left[2\pi\operatorname{Res}(f,0)\right]\end{align} where $$f(z)=\sum_{r=0}^\infty\left(\frac i2\right)^r\frac{z^{n-r-1}(1-z^2)^r}{r!}.$$ To evaluate this residue we seek the term independent of $z$ in $zf(z)$. This occurs when the power of $1-z^2$ is $(r-n)/2$ which means that $r=2s$ is even. Hence \begin{align}\operatorname{Res}(f,0)&=\sum_{s=n/2}^\infty\left(\frac i2\right)^{2m}\frac{(-1)^{(2s-n)/2}}{(2s)!}\binom{2s}{(2s-n)/2}\\&=\sum_{s=n/2}^\infty\frac{(-1)^s}{2^{2s}}\frac{(-1)^{s-n/2}}{(s-n/2)!(s+n/2)!}.\end{align} Substituting back into the integral gives $$I_n=\sum_{s=n/2}^\infty\frac{(-1)^{n/2}\pi}{2^{2s-1}(s-n/2)!(s+n/2)!}=\sum_{t=0}^\infty\frac{(-1)^{n/2}\pi}{2^{2t+n-1}t!(t+n)!}=2\pi i^n{\cal I}_n(1)$$ and we can immediately conclude that $I_{4k}>0$ and $I_{4k+2}<0$.