Settling time of sixth order denominator transfer function
I'll add a few bits besides user287001's answer (who has the answer you should be selecting) and jonk's comments. The settling time is generally accepted to be the time it takes for the response to settle within 1% of the final value. As per the answer above, the closer the real parts of the roots are to the \$j\omega\$ axis, the longer the settling time, and the lower frequency these poles have, the more they influence the output.
Consider three sets of roots, each with three complex conjugate pairs: r1 has a pole closer to the origin by a factor of around 10, compared to the others, r2 has twice that value of the real part for the first pair, and r3 has the realpart of the 3rd pair close to the \$j\omega\$ axis. The three resulting polynomials will be their denominators. For convenience, the transfer functions are for an all-pole lowpass.
r1=[-1-i, -1+i, -10-5i, -10+5i, -20-30i, -20+30i];
r2=[-1-r1(1:2) r1(3:end)];
r3=[r1(1:4) 19+r1(5:6)];
p1=poly(r1),
p2=poly(r2),
p3=poly(r3),
p1 =

        1       62     2347    35570   228950   387000   325000

p2 =

        1       64     2470    40200   297625   805000   812500

p3 =

        1       24     1112    20446   151297   261790   225250

Their values are all quite similar, with the minor exceptions that p2 has 8.125e5 instead of 3.25e5, and p3 has lower \$s^5\$ term (and onward), but similar for \$s^0\$. If you were to look at any of these, without any other comparison, would you be able to tell which one had the fastest settling time? Looking at all three, you might guess that the \$s^0\$ term for all will influence the most their corner frequency and, while true, it wouldn't be quite so true numerically:

Despite the large difference between p1 and p2, the difference is less than 20%, and that's because the last term has, in fact, the power of 6. OTOH, that doesn't tell you anything about the damping factor(s) of the forming 2nd order sections, those would influence the other terms except \$s^6\$, and it's these ones that affect the settling time:

Out of the three, p2 influences the most the settling time because the magnitude of the pole is 1.58 times greater and still closer to the origin than the others, while p2 and p3, despite an almost twice as large difference in magnitude for the last pole, their responses are similar because they are also 20 times farther from the unit circle.
So, the conclusion to take is that there is no analytical solution, and even &quot;informed guessing&quot; can't help much. Also, what you said in the comment is not applicable, because if you can simply discard the greater term, what's stopping you from discarding the next largest, and then on and on? And a 5th order has no analytical roots, a 4th does, but have you seen the horrors? This is one root for the generic \$x^4+ax^3+bx^2+cx+d\$:

I dare you to find a numerically-friendly way to calculate this (i mean in terms of numerical accuracy). You, yourself, are talking about root finding, which are complex operations -- doesn't this make you ponder about the possibility of finding simplicity in such complex operations?
Still, if the settling time is not your concern but the step response is (the time for the response to reach the 50% value), then you could use free term and the larger the value, the faster the response. As for the rising time (10% to 90%), that's also dependent on the damping(s), so that's also off.
All in all, I'd say that your quest for finding that one magical formula to determine the settling time for higher order systems is not worth it, because higher order systems are inherently complex and, thus, impossible to describe in a few words. Except, maybe, &quot;they're complex&quot;.