Busy Beavers: quadratic relation of output size and running time?
It's true, all current busy beaver candidates have a quadratic relationship between output size and steps. As Ross points out, this has to do with how they act. They move back and forth across much of the output every time they increase the output size by some constant. That means that to get to size N they must run for approximately 1 + 1 + 2 + 2 + 3 + 3 + ... = N(N+1) steps.

However, there are many other types of Turing machines. For example, there are ones often called counters which move back and forth across a section of output, each time incrementing a binary (or other base) counter. These take exponential time to increase the size of the output. One interesting one is Brady's "Surprise in a Box" http://www.drb.insel.de/~heiner/BB/simAB3Y_SB.html which expands quadratically for a little while and then moves completely within the "box" until it halts.

There are also machines which have a sort of recursive "back and forth" behavior so that it takes them N^2 steps to increase the size of the tape. They have a cubic relationship between output size and steps.

I suspect that for larger values busy beavers will not have this relation because they will get more complex in their behavior, but I don't know of any proofs.