Edited in response to revised question and additional comments by the OP.
I disagree with @JasonR's assertion that filter ringing is due to Gibbs phenomenon.
As described in the Wikipedia article linked to in Jason's answer,
the Gibbs phenomenon is an observation about the asymptotic behavior
of the truncated sum (first
$n$ terms) of the Fourier series of a periodic but discontinuous
signal such as a square wave or sawtooth wave. The Wikipedia article
illustrates an example of the square wave, showing that as more and more
terms are taken ($n$ gets large), the truncated Fourier sum becomes
closer and closer to the square wave. There are oscillations that
occur around the switching instants where the square wave transitions
from high to low or vice versa, but these become smaller and smaller
as $n$ gets large. As Jason correctly points out, the amplitude of
the oscillations becomes smaller, the frequency increases, and the
(observed) duration also becomes smaller. Overall, it looks like
the truncated Fourier
sum is converging to the square wave in the limit as $n \to \infty$.
The Gibbs phenomenon is the observation that even in the limit
as $n$ goes to $\infty$, the Fourier series sum does not
converge to the high value or the low value at the switching
instants where the square wave changes value abruptly. (Convergence
does occur at all other time instants). This has
nothing to do with filtering per se, except in the sense
that the truncated Fourier sum can be thought of as the output
of an ideal brick
wall low-pass filter with square wave input. If the
filter cut-off is such that the first $n$ harmonics are
passed through unchanged and higher harmonics are blocked,
the output is the truncated Fourier sum of the first $n$
terms. But in the limit, which is when the Gibbs phenomenon
occurs, there is no filter: all the harmonics are passed through
to the output without any change. For this reason, I do not
agree that filter ringing is due to the Gibbs phenomenon.
So why does ringing occur?
All (nontrivial) filters ring, regardless of whether they
are brick-wall or not, regardless of the shape of the input
signal, and regardless of whether the input is continuous
or has sharp transitions. The reason is that if the input
has energy in the frequency bands that are stopped (whether
wholly or in substantial part), that energy is effectively
stored internally in the filter and released slowly as
in-band energy as time progresses.
Most of the time this release is not noticed very much
because it is drowned out by the response to the in-band
signal that is present. However, if the in-band
signal changes (or ceases) relatively suddenly, that energy stored
from previous times still has to be released, and this is the ringing
that is observed after the in-band signal has disappeared.
In DSP terms, the FIR filter buffer continues to empty out
even after the
signal has ended, and so the output continues even after the signal ends.
Since sharp-cutoff filters have long buffers
(many biquad sections if you will), this emptying takes a long time
and is much more noticeable than with a more easy-going filter
which empties out quite quickly.