Here’s a complete solution.
It’s convenient to represent each new number as the ordered pair of its left and right parents. The first new number is $2$, represented by $\langle 1,1\rangle$. The next two are both $3$, represented by $\langle 1,2\rangle$ and $\langle 2,1\rangle$. At the next step there are four new numbers, $4,5,5,4$, represented respectively by $\langle 1,3\rangle$, $\langle 3,2\rangle$, $\langle 2,3\rangle$, and $\langle 3,1\rangle$. Of course each new number is simply the sum of its parents, so $\langle m,n\rangle$ always represents $m+n$. It’s also clear that $\langle m,n\rangle$ is the right parent of $\langle m,m+n\rangle$ and the left parent of $\langle m+n,n\rangle$. Let’s follow the descendants of $\langle m,n\rangle$ and their cubes for a few generations.
From $\langle m,n\rangle$ we get $\langle m,m+n\rangle$ and $\langle m+n,n\rangle$, corresponding to $2m+n$ and $m+2n$, the sum of whose cubes is $9(m^3+2m^2n+2mn^2+n^3)$. These produce $\langle m,2m+n\rangle$, $\langle 2m+n,m+n\rangle$, $\langle m+n,m+2n\rangle$, and $\langle m+2n,n\rangle$, corresponding to $3m+n$, $3m+2n$, $2m+3n$, and $m+3n$, the sum of whose cubes is $7\cdot9(m^3+2m^2n+2mn^2+n^3)$.
If we call $\langle 1,1\rangle$ the first generation, $\langle 1,2\rangle$ and $\langle 2,1\rangle$ the second generation, and so on, the calculation in the preceding paragraph implies that for $k\ge 2$, the sum of the cubes of the numbers in the $(k+1)$-st generation is $7$ times the sum of the cubes of the numbers in the $k$-th generation.
The sums of the cubes in the first two generations are $8$ and $54$, so for $k\ge 2$ the sum of the cubes in the $k$-the generation is $s_k = 54\cdot7^{k-2}$. Recall, however, that these totals are the sums of the new cubes at each stage of the original problem. Let $t_k$ by the sum of the cubes at stage $k$ in the original problem. Then $t_{k+1}=t_k+s_k$ for $k\ge 1$, where $t_1=2$. Thus, $t_1=2$, $t_2=2+8=10$, and $t_{k+1}=t_k+s_k=t_k+54\cdot7^{k-2}$ for $k\ge 2$. In other words, for $k\ge 2$ we have
$\begin{align*} t_{k+1}&=t_2+\sum_{i=0}^{k-2}\left(54\cdot7^i\right)\\ &=10+54\cdot\frac{7^{k-1}-1}6\\ &=10+9\left(7^{k-1}-1\right)\\ &=1+9\cdot7^{k-1}\;. \end{align*}$
This formula also yields the correct value when $k=1$, so after shifting the index we have $t_k=1+9\cdot7^{k-2}$ for all $k\ge 2$, as desired.