Are overdamped rise waveforms really "curvier"? Or is that just time scale?
I was thinking I would have to get my pole placement just right to get
  about 75° phase shift at exactly the crossover frequency, even if I
  could have say 85° at crossover


For a simple 2nd order low pass filter here's what the phase shift is (bottom left): -



Unless you are envisaging a more complex filter, the phase response (\$\omega_{natural}\$ normalized to 1) is always going to be -90 degrees. And, of course for this example \$\omega_{natural}\$ = \$\dfrac{1}{\sqrt{LC}}\$

The amplitude response i.e. how peaky the spectrum is (top left picture) is always equal to the Q of the circuit.

So as damping increases (Q decreases) the maths becomes less that of a tuned resonant (and ringing) formula to that of an RC (exponential) circuit. In other words (in this example), L starts to be swamped by the dominance of R.

The upshot of this is that at damping ratios of 1 and 2 (but with a higher natural resonant frequency for the higher damping ratio), the two responses will "overlay" and look quite similar BUT, they won't be the same. For damping ratios of 10 and 20 (with the appropriate increase of resonant frequency for the damping ratio of 20), the responses will look extremely similar because of the dominance of R (over L) in both these cases.

It becomes just an RC circuit and therefore, providing the product of R and C are equal, the response will be progressively more similar as damping increases.