Iterations of the radical of an integer We denote for an integer $n>1$ its square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,\tag{0}$$
with the definition $\operatorname{rad}(1)=1$. You can see this definition and the properties of this arithmetic function from the Wikipedia Radical of an integer. Then I wondered about a variant of two problems showed in last paragraphs of section B 41 of Guy's book [1].
Definition. For each fixed integer $n\geq 1$, I consider the arithmetic function $$f(n)=f^1(n):=\operatorname{rad}(n)+1,\tag{1}$$
and we create the iterated $$f^{k+1}=f(f^{k}(n))\tag{2}$$
until $f^{T}(n)$ reaches a prime number, here thus $T=T(n)$ is a function depending on $n$ (that is the number of steps in our iteration that we need to wait until the sequence of iterated $f^{k}(n)$ reaches a prime number 
for the first time). We can call stopping time to the arithmetic function $T(n)$ (defined thus for integers $n\geq 1$ and taking positive values).
Example 1. Let $n=1$, then in a step the sequence of iterations reaches a prime number since $f(1)=f^1(1)=\operatorname{rad}(1)+1=1+1=2$ is a prime number. The same argument works with the composite number $n=8$, because $f(8)=f^1(8)=\operatorname{rad}(8)+1=2+1=3$ a prime number. Thus in our table $T(1)=1$ iteration, and also in this example we've seen that the corresponding  sequence defined for the integer $8$ has stopping time $T(8)=1$.
Example 2. We work about the calculation of $T(33)$. We iterate while the result is a composite integer. Here $\operatorname{rad}(33)=33$, thus $f(33)=34$ and since $34=2\cdot 17$ is also square-free then $f^2(33)=f(f(33))=34+1=35$. Again our last iterated, that is $35=5\cdot 7$ has no repeated prime factors, thus we write $f^3(33)=f(35)=36$. Finally $f^4(33)=\operatorname{rad}(36)+1=6+1=7$ that results a prime number. Thus we conclude that $T(33)=4$.
Computational fact. I did few experiments/calculations and seems that the arithmetic function $T(n)$ is erratic, thus it makes sense to study averages of consecutive values of this stopping time.

Question. I would like to know if it is possible to deduce a tighter upper bound for $$\frac{1}{N}\sum_{1\leq n\leq N}T(n)\tag{3}$$
  as $N$ grows (if you are able to provide an elaborated statement about the asymptotic behaviour of $(3)$ as $N\to\infty$ feel free to do it). Many thanks.

Final remarks. My motivation was to propose a similar variant of the mentioned problems in Guy's book, now for other interesting arithmetic function $\operatorname{rad}(n)$. I've no idea about a strategy to attack the problem to get idea about the size of the average means of the $T(n)$.
References:
[1]  Richard K. Guy, Unsolved Problems in Number Theory, Volume I, Second Edition, Springer (1994).
 A: Based on some computations:
I would say that $ln(n)$ (the red line) appears to be a good, safe upper bound to your interesting averaged-$T(n)$ (the blue line).
I shall try to think of some ways of using bounds on $rad(n)$ or $\pi(x)$ to obtain this result, but I'm not terribly good at this sort of thing (just very, very interested!). It looks to be roughly a logarithmic growth itself.
EDIT: I have just messed around a bit with the logarithms, and here's what a $log_{24}(n)$ graph looks like:
Could be a good lower bound beyond the intersection point (roughly $x=217$)?
Thus the current working conjecture is $log_{24}(x)<\frac{1}{x}\sum^x_{k=1} T(k)<log_e(x)$ beyond around $x=220$.
EDIT II: Considering the information gained from the other answer, we can derive some nice asymptotic upper bounds on the average-$T(n)$.
We see from the fact that $T(n) \leq \frac{4}{ln(2)}ln(n)$ that $\frac{1}{n}\sum_{1\leq k \leq n}T(n) \leq \frac{4}{n ln(2)}\sum ln(k) = \frac{4}{n ln(2)} \times ln(n!)$. 
Now, Stirling's Formula tells us that $ln(n!)$ is asymptotic to $ln(\sqrt{2\pi n})+nln(\frac{n}{e}) = ln(\sqrt{2\pi n})+ln(n)-n$.
So, putting this all together, we get that the average-$T(n)$ should have the upper bound of $$\frac{4(ln(\sqrt{2\pi n})+ln(n)-n)}{n ln(2)}$$ Pretty cool!
A: Note that $f(n)=n+1$ if $n$ is square-free, which can happen at most three times in a row when iterating $f$, namely one of $n,n+1,n+2,n+3$ must be a multiple of $4$. And if $n$ is not square-free, then $2\le f(n)\le \frac n2+1$. Thus it takes up to four iterations to take $n$ to a number $\le \frac{n+3}2+1$. Thus if we start with $n\le 5+2^m$, we arrive at a number $\le 5+2^{m-1}$ after at most $4$ iterations.
Thus after at most $4\log_2(n-5)$ we are at a prime or at $4$ or $6$.
This gives the (correct, but probably very pessimistic) upper bound
$$ T(n)\le 1+4\log_2(n-5)\sim \frac 4{\ln2}\cdot \ln n\approx 5.77\ln n$$
(which then holds in similar form for the averaged $T$).

A more heuristic appoach is that the probability of $n$ beinf squarefree is $\frac 6{\pi^2}$, hence on average, we should be able to replace the factor $4$ above with $\frac{\pi^2}6\approx1.64$. We might also note that when $n$ is a multiple of $4$, its is a multiple of $8$ or higher about half the time etc. Thus we should then not step from $\sim 2^m$ to $\sim 2^{m-1}$, but on average more like to $2^{m-1-\frac12-\frac14-\ldots}=2^{m-2}$. Boldly combining both ideas, we get a heuristic estimate of $\sim \frac{\pi^2}6\log_4n\approx 1.19\ln n$. And there is still room for impovement by stopping t other primes than $2,3,5,7$.
