The numbers $p_i$ in your list are of the form $p_i=344554\cdot2^{7-i}3^i-1,$ for $i=0,1,\ldots,7$, so there is nothing mysterious about the appearance of a row from Nicomachus' triangle, as $ \frac{p_{i+1}-p_i}{344554}=3^{i+1}2^{6-i}-3^i2^{7-i}=3^i2^{6-i}(3-2)=3^i2^{6-i}. $ Of course, it is rare that they would all primes, but as Gerry Myerson explained, there is no obvious reason, why we couldn't find even longer sequences of primes given that the residue class of $-1$ is a fixed point for the mapping $x\mapsto(3x+1)/2$ modulo $q$, $q$ any odd prime.
As an example of a constraint to your search for such sequences, implicit in Gerry's answer, let us consider the following. If you search for such a sequence of length ten, then all the primes in that sequence must be congruent to $-1$ modulo $11$. This is because the mapping $x\mapsto(3x+1)/2$ permutes the other residue classes modulo $11$ as a 10-cycle: $ 6\mapsto 4\mapsto 1\mapsto 2\mapsto 9\mapsto 3\mapsto 5\mapsto8\mapsto 7\mapsto 0(\mapsto 6), $ so any integer $x$ that is not $\equiv-1\pmod{11}$ will become divisible by eleven after at most nine iterations.
Trying to address the question on the emergence of Nicomachus' triangle. Assume that we start a Collatz run from the number $a_0=m\cdot 2^k-1$, where $k>0$ and $m$ is an odd integer. Observe that every odd positive integer $a_0$ can be written in this way. As $a_0$ is odd, the next Collatz step gives $3a_0+1=m\cdot2^k\cdot3-2$. This is an even number, so the next step yields $a_1=m\cdot2^{k-1}\cdot3-1.$ If $k>1$, then we can repeat the same reasoning, and two further steps of Collatz gives us $a_2=(3a_1+1)/2=m\cdot2^{k-2}\cdot3^2-1.$ The logic will be broken after $2k$ Collatz steps, because $a_k=m\cdot2^03^k-1$ is an even number, so the two types of steps stop alternating at this point.
Assume that the numbers $a_i$ are primes in the range $i=0,1,\ldots,\ell$. Obviously then $\ell. Let us consider the differences $\Delta_i=a_{i}-a_{i-1}$, for $i=1,2,\ldots,\ell$. Then $ \Delta_i=(m\cdot2^{k-i}3^i-1)-(m\cdot2^{k-i+1}3^{i-1}-1)=m\cdot2^{k-i}3^{i-1}(3-2) =m\cdot2^{k-i}3^{i-1}. $ We immediately see that the greatest common divisor of the numbers $\Delta_i,i=1,2,\ldots,\ell$ is $d=m\cdot2^{k-\ell}.$ This is because, we always have $3\Delta_i=2\Delta_{i+1}$. Thus the numbers $ \frac{\Delta_i}{d}=\frac{m\cdot2^{k-i}3^{i-1}}{m\cdot2^{k-\ell}}=2^{\ell-i}3^{i-1} $ form the $\ell^{\text{th}}$ row Nicomachus' triangle with $i$ ranging from $1$ to $\ell$.