if a standard three-phase 400V AC connection is rectified what DC voltage comes out of it?
If you measure across the load shown in your figure, the peak voltage will be ~565V; the DC voltage will depend on your load and filtering as others have noted.

If you measure from the + of the load in your figure to the neutral of your AC supply, the peak voltage will be ~325V. If you hook up a load like that, you aren't actually using a full-wave rectifier.

The simplest way to get 565V is to start from 400V and apply the standard \$\sqrt{2}\$ scaling from \$V_{rms}\$ to \$V_{p-p}\$. However, starting from 400V is skipping part of the calculation. The more thorough way to derive 565V is to calculate it as:

$$(325 \text{V}) * \max_{\theta} \left\{ \sin (\theta + \frac{2 \pi}{3}) - \sin (\theta)  \right\}$$

The expression is maximized when \$\theta\$ is \$\frac{5 \pi}{3}\$, and the maximum value is \$325 \sqrt{3} = 563\$.

There is a detailed analysis including some java applets here.