It is not a complex analytic technique but I think it is worth mentioning. We can compute the integral by taking Riemann sums and exploiting the identity: $$\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}\tag{1}$$ from which it follows that: $$\begin{eqnarray*}\int_{0}^{1}\log(\sin(\pi x))\,dx &=& \frac{1}{\pi}\int_{0}^{\pi}\log\sin x\,dx = \frac{1}{\pi}\lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}\\&=&\lim_{n\to +\infty}\frac{1}{n}\,\log\frac{2n}{2^n}=\color{red}{-\log 2}.\tag{2}\end{eqnarray*}$$

1. Riemann sums

It is not a complex analytic technique but I think it is worth mentioning. We can compute the integral by taking Riemann sums and exploiting the identity: $$\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}\tag{1}$$ from which it follows that: $$\begin{eqnarray*}\int_{0}^{1}\log(\sin(\pi x))\,dx &=& \frac{1}{\pi}\int_{0}^{\pi}\log\sin x\,dx = \frac{1}{\pi}\lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}\\&=&\lim_{n\to +\infty}\frac{1}{n}\,\log\frac{2n}{2^n}=\color{red}{-\log 2}.\tag{2}\end{eqnarray*}$$

Other approaches deserve to be mentioned:

2. Symmetry

The function $\sin(\pi x)$ is symmetric with respect to $x=\frac{1}{2}$, hence $$\begin{eqnarray*}I=\int_{0}^{1}\log\sin(\pi x)\,dx&\stackrel{x\to 2z}{=}&2\int_{0}^{1/2}\left[\log(2)+\log\sin(\pi z)+\log\cos(\pi z)\right]\,dz\\&=&\log(2)+2I.\end{eqnarray*}\tag{3}$$

3. An obscene overkill

By Raabe's theorem $\int_{a}^{a+1}\log\Gamma(x)\,dx = \log\sqrt{2\pi}+a\log a-a$ and by the reflection formula for the $\Gamma$ function $\frac{\pi}{\sin(\pi z)}=\Gamma(z)\Gamma(1-z)$, hence the question is trivial by switching to logarithms and integrating over $(0,1)$.