Here is an outline:
- No matter what $N$ you start with, it is impossible to end with $3$ or higher. The penultimate $N$-value would have either been $N=1$ (impossible, since your next step would need to have been subtracting 2) or $N=5$. Since $N$ was $5$ at some point, the prime $3$ was in the original set of primes, and this necessitates an even earlier step in the chain. The step before that would have either been with $N=2$ (again impossible, since you would have subtracted 3) or with $N=8$. And again, we have a number so large that there had to have been another prime in the original list ($5$); there had to be an earlier step; and there had to have been a point when $N$ was much larger ($13$). Since the value of $N$ is getting larger at a more than quadratic rate (summing primes grows almost like like $n^2\log(n)$) while prime values are getting larger much less quickly, this process will never have a chance to end.
- No matter what $N$ you start with, it is impossible to end with $-2$ or lower. A similar, but negative, argument applies. If we end with $-2$, then the penultimate value was either $N=0$ (not possible, since our last step would then have been adding $2$) or $N=-5$. Since $N$ is negative at this step, there must have been a previous step, where either $N=-2$ (again not possible) or $N=-8$. And this continues; we never return to an initial $N$-value that is positive, since $N$ is growing negative so quickly relative to the next largest prime.
- Since you are subtracting and adding primes (which are mostly odd) and you are starting with $N$ prime (and odd), the parity of the terminal number only depends on how many steps there are in your sequence - how many primes there are up to $N$. For example, since $N=11$ yields a prime collection $\{2,3,5,7,11\}$ with 5 elements, you will add/subtract 4 odd numbers, and end up with an odd terminal $N$-value. But with $N=13$, you will add/subtract 5 odd numbers, and end up with an even terminal $N$-value.
This explains why the terminal values of $N$ would have to alternate between something in $\{-1,1\}$ and something in $\{0,2\}$ as $N$ increments through prime values.
Your conjecture would be proved if we could rule out $-1$ as a terminal value when $N$ is prime. I've tried assuming that $-1$ is a terminal value, hoping that this led to only finitely many possible initial $N$, none of which are prime. However it appears that many $N$ lead to $-1$. The smallest are: $4, 10, 16, 22$. Perhaps it can be show that if $-1$ is terminal for $N$, then $N$ was even. But maybe someone else can take it from here.
Hope this much helps!
EDIT added MUCH later
We can indeed show that if the terminal value is $-1$, then $N$ must have been even.
Working again from the end of the process backward, let's denote $p_i$ to be the $i$th prime in the list, and $Z_i$ the $i$th value for $Z$. Again, these indices are from the end of the process. So we take $Z_0=-1$ and $p_0=2$.
$p_1$ is $3$. And $Z_1$ must have either been $-1+p_0$ or $-1-p_0$; that is, either $1$ or $-3$. Note that the only possibilities for $Z_1$ are odd numbers.
Next, $p_2$ is 5, and $Z_2$ must be one of the preceding possibilities for $Z_1$ plus or minus $3$. So $Z_2$ must be even.
Since all further primes are odd, if we continue in this way we see that for $n>0$, whatever $Z_n$ is, it's odd if $n$ is odd and even if $n$ is even.
Now only for some of the potential $Z_n$ that we can reverse engineer this way, is it possible for them to have been initial $N$ values. Still, we know now that if we start with an $N$ that has $-1$ as its terminal $Z$-value, and if it takes an even number of steps to get there, then $N$ must have been even, and therefore not prime.
All told, this process has terminating $Z$-values that alternate between $1$ and either $0$ or $2$ as $N$ increments through prime values.
Best Answer
The 'odds' that a random odd number $\lt 10^7$ is prime are roughly $2/\ln(10^7)\approx1/8$, which comes pretty close to your 11.8% figure. I suspect there's nothing known about this but I would expect them to follow the heuristics. If I'm doing my pre-coffee math correctly, then since abstractly the number you're hoping to be prime is of size $\approx p^2$, heuristically it seems as though chains of arbitrary fixed length should exist: for all $p$, the heuristic probability of a chain of length $n$ starting from $p$ is $\displaystyle\approx \frac1{\ln p}\cdot\frac1{\ln(p^2)}\cdots\frac1{\ln(p^{2^n})} \approx \frac1{2^{n^2/2}\ln^np}$. The constant term is smaller, but this is broadly akin to the probability for Cunningham Chains of length $n$.