> n independent random steps will, on the average, end up about sqrt(n) steps from the origin

AKCSHUALLY

The root of the mean of the squared distances is sqrt(n).

The phrase as you quoted is the mean of the absolute distance. In general those are different. I don't know the latter from memory, and a quick look at the wikipedia page for Brownian motion doesn't have it.

You're right, sqrt(E[d^2]) > E[|d|] because of Jensen's inequality. The latter is about 0.8 sqrt(N).