I think it's intuitive to look at it in the time domain. After all, ringing and over- or undershoot are time-domain phenomena. Of course, damping is an important factor, but looking at the impulse response can give you an immediate idea whether there will be ringing and/or overshoot in the step response.

Note that the step response $s(t)$ is the integral of the impulse response $h(t)$:

$$s(t)=\int_{-\infty}^th(\tau)d\tau\tag{1}$$

So if there are oscillations in the impulse response, there will be ringing in the step response. Of course, low damping (high Q) will result in larger oscillations which are decaying more slowly, hence more ringing.

If there are no oscillations at all, i.e., if the impulse response is non-negative, there will be no ringing. A Gaussian impulse response has this property. Impulse responses of Bessel filters are similar to Gaussians, and they approach a Gaussian with increasing order. Consequently, the step response of Bessel filters exhibits very little ringing.

The flat group delay of Bessel filters is only indirectly related to the relative absence of ringing. What is more important is the relatively soft decay of the magnitude of the frequency response (which is caused by the side constraints of flat group delay). However, we can also have zero ringing with a completely non-linear phase, as long as the impulse response is non-negative. And, vice versa, we can find linear phase filters with a lot of ringing. The ringing is caused by the filter's frequency selectivity, and the sharper the transitions from passband to stopband, the more ringing will occur.

I think it's intuitive to look at it in the time domain. After all, ringing and over- or undershoot are time-domain phenomena. Of course, damping is an important factor, but looking at the impulse response can give you an immediate idea whether there will be ringing and/or overshoot in the step response.

Note that the step response $s(t)$ is the integral of the impulse response $h(t)$:

$$s(t)=\int_{-\infty}^th(\tau)d\tau\tag{1}$$

So if there are oscillations in the impulse response, there will be ringing in the step response. Of course, low damping (high Q) will result in larger oscillations which are decaying more slowly, hence more ringing.

If there are no oscillations at all, i.e., if the impulse response is non-negative, there will be no ringing. A Gaussian impulse response has this property. Impulse responses of Bessel filters are similar to Gaussians, and they approach a Gaussian with increasing order. Consequently, the step response of Bessel filters exhibits very little ringing.

The flat group delay of Bessel filters is only indirectly related to the relative absence of ringing. What is more important is the relatively soft decay of the magnitude of the frequency response (which in turn is a consequence of the side constraints of a flat group delay). However, we can also have zero ringing with a completely non-linear phase, as long as the impulse response is non-negative. And, vice versa, we can find linear phase filters with a lot of ringing. The ringing is caused by the filter's frequency selectivity, and the sharper the transitions from passband to stopband, the more ringing will occur.