Here a start:

We have the [reflection formula][1]
$$z! (1-z)! = \frac{\pi z (1-z)}{\sin (\pi z)}.$$
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?



  [1]: http://en.wikipedia.org/wiki/Gamma_function#Properties