Conjecture:
Let $f(x,y,z)=xy+xz+yz\:$ and
$\: S_m=\{n\in\mathbb N|n<m\wedge f(p_n,p_{n+1},p_{n+2})\in\mathbb P\}$.
Then $|S_m|\sim \frac{m}{\ln m}$.
Computationally tested only, but I would like to see a proof.
It seems like I made a mistake with the inequality in the comment, but here are some data:
m m/ln m |Sm| pi(m) li m |Tm|
10 4.34 5 4 6.1655995 6
100 21.71 26 25 30.1261416 41
1000 144.76 171 168 177.6096580 295
10000 1085.74 1292 1229 1246.1372159 2234
100000 8685.89 10102 9592 9629.8090011 12024
1000000 72382.41 82114 78794 78627.5491595
5000000 324150.19 366716 348513 348638.1150413
For comparision: $T_m=\{n\in\mathbb N|n<m\wedge f(a_n,a_{n+1},a_{n+2})\in\mathbb P\}$, where $a_n=2n+1$
There appears to be little hope of a formal proof, but this is how a standard heuristic argument would treat this problem, so we can determine whether the conjecture is plausible in the vein of Bateman-Horn as mentioned by reuns.
Excluding an asymptotically small number of cases, most of the values $f(p_n,p_{n+1},p_{n+2})$ will be of size roughly $\Theta(m^2 \log^2 m)$, so if they were randomly distributed, we would expect about one in $\log(m^2\log^2 m) \sim 2 \log m$ of them to be prime, which would be $\sim m/(2 \log m)$.
But we're seeing significantly more than that. That's because those numbers are not randomly distributed: for starters, they're all odd, and that accounts for the missing factor of $2$. A random number has a $\frac12$ chance of not being divisible by $2$, but these numbers have double the chance.
However, we're not done yet. They also are unevenly distributed modulo $3$: each of $p_n,p_{n+1},p_{n+2}$ can be viewed as a uniform random sample from the non-zero residues mod $3$. Of these $8$ possible inputs to $xy+yz+zx$, $2$ of them provide $0$. So there is a $\frac34$ chance of these numbers being indivisible by $3$, versus a $\frac23$ chance for a random number: which is a factor of $\frac98$ better. Not as strong as the previous factor of $2$, but not insignificant.
It turns out we can exactly calculate this factor for any given prime modulus $p$: the number of solutions to $xy+yz+zx \equiv 0 \pmod p$ with $x,y,z$ all non-zero is easy to compute, since it's equivalent to $z(x+y) \equiv -xy$. A straightforward calculation gives $(p-1)(p-2)$ zero-producing combinations out of $(p-1)^3$, for a $\frac{p-2}{(p-1)^2}$ chance of being divisible by $p$, which is an advantage of $\frac{p(p^2-3p+3)}{(p-1)^3}$.
For large $p$ this fraction is approximately $1 + \frac{1}{p^2}$, so we can combine all of these multiplicative factors for all primes $p$ into a convergent product yielding a single constant: this is the so-called singular series.
The standard heuristic underlying Bateman-Horn and similar conjectures would then predict that the number of prime values should be asymptotic to:
$$\frac{m}{2\log m}\prod_{p\text{ prime}} \frac{p(p^2-3p+3)}{(p-1)^3} \approx 1.15048 \frac{m}{\log m}.$$
This actually agrees remarkably well with your data, which shows the ratios of $|S_m|$ to $m/\log m$ are:
$$1.1521, 1.1976, 1.1813, 1.1900, 1.1630, 1.1344, 1.1313.$$
However, I don't have a good explanation for why it matches so well, since I would have expected a $O(1/\log m)$ relative error so for these small values of $m$ it should be agreeing to only one digit, not two. There may be something that miraculously cancels out the first-order error and leaves a $O(1/\log^2 m)$ relative error.
Update: For the similar sequence $T_m$, we can do a similar analysis with the polynomial $3(2n+1)^2-4$. The singular series constant (with the factor of $\frac12$ due to the quadratic growth rate) works out $$\frac12 \cdot 2 \cdot \frac32 \cdot \prod_{p\ge 5\text{ prime}}\big(1-\frac1p\big)^{-1}\big(1-\frac{(3/p)}{p}\big) \approx 2.07508,$$ where $(3/p)$ is the Legendre symbol (so $+1$ if $p \equiv 1,11$ mod $12$ and $-1$ if $p \equiv 5,7$ mod $12$).
This is significantly higher than the constant for $S_m$: much of the reason for this is that $3(2n+1)^2-4$ is lucky enough to be never divisible by any of the first four primes $2,3,5,7$ so it tends to be prime more often than a random number of the same size. The factor of about $2.07$ seems to agree very well with the first few rows of your table, so you may want to double-check the computation of $T_{100000}$ as it seems not to fit the curve.