\begin{align*}
&\sum_{n = -\infty}^{\infty} \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} \\
&= \frac{\log \left( (\frac{1}{3})^2 \right)}{\frac{1}{3}} + \sum_{n = 1}^{\infty} \left( \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} + \frac{\log \left( (-n + \frac{1}{3})^2 \right)}{-n + \frac{1}{3}} \right) \\
&= -3\log(9) + \sum_{n = 1}^{\infty} \left( \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} - \frac{\log \left( (n - \frac{1}{3})^2 \right)}{n - \frac{1}{3}} \right) \\
&= -3\log(9) + 6\sum_{n = 1}^{\infty} \left( \frac{\log \left( \frac{3n + 1}{3} \right)}{3n + 1} - \frac{\log \left( \frac{3n - 1}{3} \right)}{3n - 1} \right) \\
&= -6\log(3) + 6\sum_{n = 1}^{\infty} \left( \frac{\log(3)}{3n - 1} - \frac{\log(3)}{3n + 1} \right) \\
&\quad+ 6\sum_{n = 1}^{\infty} \left( \frac{\log \left( 3n + 1 \right)}{3n + 1} - \frac{\log \left( 3n - 1 \right)}{3n - 1} \right) \\
&= -6\log(3) + 6\sum_{n = 1}^{\infty} \left( \frac{\log(3)}{3n - 1} - \frac{\log(3)}{3n + 1} \right) \\
&\quad+ 6\sum_{n = 1}^{\infty} \left( \frac{\log \left( 3n + 1 \right)}{3n + 1} - \frac{\log \left( 3n - 1 \right)}{3n - 1} \right)
\end{align*}
But
\begin{align*}
&-6\log(3) + 6\sum_{n = 1}^{\infty} \left( \frac{\log(3)}{3n - 1} - \frac{\log(3)}{3n + 1} \right) \\
&= -6\log(3)\left(1 + \sum_{n = 1}^{\infty} \left( \frac{1}{3n + 1} - \frac{1}{3n - 1} \right) \right) \\
&= -6\log(3)\frac{2}{\sqrt{3}}\sum_{n = 1}^{\infty} \frac{\sin \frac{2n\pi}{3}}{n} \\
&= -\frac{4\sqrt{3}\log(3)}{2i}\sum_{n = 1}^{\infty} \frac{e^{2i\pi n/3} - e^{-2i\pi n/3}}{n} \\
&= -\frac{4\sqrt{3}\log(3)}{2i}\log\frac{1 - e^{-2i\pi/3}}{1 - e^{2i\pi/3}} \\
&= -\frac{4\sqrt{3}\log(3)}{2i}\log\left(-e^{e^{-2i\pi/3}}\right) \\
&= -\frac{4\sqrt{3}\log(3)}{2i}\log(e^{i\pi/3}) \\
&= -\frac{4\sqrt{3}\log(3)}{2i}\left(i\frac{\pi}{3}\right) \\
&= -\frac{2\pi\log(3)}{\sqrt{3}}
\end{align*}
So
\begin{align*}
&\sum_{n = -\infty}^{\infty} \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} \\
&= -\frac{2\pi\log(3)}{\sqrt{3}} + 6\sum_{n = 1}^{\infty} \left( \frac{\log \left( 3n + 1 \right)}{3n + 1} - \frac{\log \left( 3n - 1 \right)}{3n - 1} \right) \\
&= -\frac{2\pi\log(3)}{\sqrt{3}} + 4\sqrt{3}\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin \frac{2n\pi}{3}}{n}
\end{align*}
From Kummer's formula [^1] we have
\begin{align*}
&\log \Gamma(x) = \frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin(2\pi nx)}{n} \\
&\quad+ \frac{\log(2\pi) + \gamma}{\pi}\sum_{n = 1}^{\infty} \frac{\sin(2\pi nx)}{n} + \frac{1}{2}\sum_{n = 1}^{\infty} \frac{\cos(2\pi nx)}{n} + \frac{1}{2}\log(2\pi)
\end{align*}
So for $x = \frac{1}{3}$ follows
\begin{align*}
&\log \Gamma\left(\frac{1}{3}\right) \\
&= \frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin\left(\frac{2\pi n}{3}\right)}{n} \\
&\quad+ \frac{\log(2\pi) + \gamma}{\pi}\sum_{n = 1}^{\infty} \frac{\sin\left(\frac{2\pi n}{3}\right)}{n} + \frac{1}{2}\sum_{n = 1}^{\infty} \frac{\cos\left(\frac{2\pi n}{3}\right)}{n} + \frac{1}{2}\log(2\pi)
\end{align*}
It holds
$$\sum_{n = 1}^{\infty} \frac{\sin nx}{n} = \frac{\pi - x}{2}$$, therefore
$$\sum_{n = 1}^{\infty} \frac{\sin\left(\frac{2\pi n}{3}\right)}{n} = \frac{\pi - 2\pi/3}{2} = \frac{\pi}{6}$$
as well as
$$\sum_{n = 1}^{\infty} \frac{\cos nx}{n} = -\log\left(2\left|\sin\frac{x}{2}\right|\right)$$, i.e.
$$\sum_{n = 1}^{\infty} \frac{\cos\left(\frac{2\pi n}{3}\right)}{n} = -\log\left(2\sin\frac{\pi}{3}\right) = -\frac{1}{2}\log 3$$
the above two sums have been proved in Mathematica
so
\begin{align*}
&\log \Gamma\left(\frac{1}{3}\right) \\
&= \frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin\left(\frac{2\pi n}{3}\right)}{n} \\
&\quad+ \frac{\log(2\pi) + \gamma}{\pi}\frac{\pi}{6} - \frac{1}{4}\log 3 + \frac{1}{2}\log(2\pi)
\end{align*}
i.e.
\begin{align*}
&\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin\left(\frac{2\pi n}{3}\right)}{n} \\
&= \frac{\pi}{12}\left(12\log\Gamma\left(\frac{1}{3}\right) - 2\gamma + 3\log 3 - 8\log(2\pi)\right)
\end{align*}
Finally
\begin{align*}
&\sum_{n = -\infty}^{\infty} \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} \\
&= \frac{\pi}{\sqrt{3}}\left(12\log\left(\Gamma\left(\frac{1}{3}\right)\right) + \log 3 - 2\gamma - 8\log(2\pi)\right)
\end{align*}