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Here a start:

We have the reflection formula $$z! (1-z)! = \frac{\pi z (1-z)}{\sin (\pi z)}.$$ Taking $\log$'s, $$\log (z!) + \log ((1-z)!) = \log \pi + \log z + \log (1-z) - \log \sin (\pi z).$$ Split our integral in half and rearrange it $$\int_0^1 \log (z!) \ dz = \int_0^{1/2} \left( \log (z!) + \log ((1-z)!) \right) dz.$$

So we have three elementary integrals to deal with, plus $$\int_0^{1/2} \log \sin(\pi z) \ dz. \quad (*)$$ According to Mathematica, $(*) = - \log(2)/2$. So, if we can find a clean proof of this fact, we will have evaluated the integral. This may be difficult, because the indefinite integral $\int \log \sin(\pi z) \ dz$ involves dilogarithms. To me, $(*)$ looks like a good target for residues. Anyone want to finish it off?

Edit: Here is a way to calculate $(*)$: Denote $I=\int_{0}^{\pi/2}\log(\sin x) \ dx=\int_{0}^{\pi/2}\log(\cos x) \ dx$ and so $$2I=\int_{0}^{\pi/2}\log(\frac{\sin 2x}{2}) \ dx=\int_{0}^{\pi/2}\log(\sin 2x) \ dx-\frac{\pi \log 2}{2}=I-\frac{\pi \log 2}{2}$$

Here a start:

We have the reflection formula $$z! (1-z)! = \frac{\pi z (1-z)}{\sin (\pi z)}.$$ Taking $\log$'s, $$\log (z!) + \log ((1-z)!) = \log \pi + \log z + \log (1-z) - \log \sin (\pi z).$$ Split our integral in half and rearrange it $$\int_0^1 \log (z!) \ dz = \int_0^{1/2} \left( \log (z!) + \log ((1-z)!) \right) dz.$$

So we have three elementary integrals to deal with, plus $$\int_0^{1/2} \log \sin(\pi z) \ dz. \quad (*)$$ According to Mathematica, $(*) = - \log(2)/2$. So, if we can find a clean proof of this fact, we will have evaluated the integral. This may be difficult, because the indefinite integral $\int \log \sin(\pi z) \ dz$ involves dilogarithms. To me, $(*)$ looks like a good target for residues. Anyone want to finish it off?