Posting my first Math-StackExchange answer...
According to this section of Wikipedia, due to Dixon's theorem, the probability of largest prime factor of $n$ to be less than $n^{1/m}$ is approximately $m^{-m}$ for any real $m \ge 1$.
So probability of largest prime factor to be less than $\sqrt n = n^{1/2}$ is approximately $2^{-2} = 1/4 = 0.25$. To be less than $\sqrt[3]n = n^{1/3}$ is approximately $3^{-3} = 1/27 \approx 0.037$.
I don't know the details of this theorem, I just found this quotation in Wikipedia and thought it might be useful for you. Also I don't know how approximate is this formula.
I tried to check this formula experimentally and wrote Python code for that (using Pollard-Rho and Fermat algorithms). Don't know if according to rules it is allowed to post code here on Math-StackExchange, so providing just links:
You can see (run) my code in action here (and here is a copy of my code just in case if first link is broken, second link is not runnable).
Results for 10K checked 64-bit numbers here:
Checked nums: 10242
Expected: 1.0: 1.00000, 1.5: 0.54433, 2.0: 0.25000, 2.5: 0.10119, 3.0: 0.03704, 3.5: 0.01247, 4.0: 0.00391, 4.5: 0.00115, 5.0: 0.00032, 5.5: 0.00008
Actual: 1.0: 1.00000, 1.5: 0.51541, 2.0: 0.23203, 2.5: 0.09008, 3.0: 0.03202, 3.5: 0.01021, 4.0: 0.00321, 4.5: 0.00105, 5.0: 0.00038, 5.5: 0.00005
Here are pairs of m
(from formula above) and probability. So Expected (by formula above) is close to Actual (experimental with factoring of 64-bit numbers), especially close for larger m
. Maybe formula is more precise for larger than 64-bit numbers that I checked or for more amount of tested numbers.