Your formula is indeed correct, and something similar does hold for similarly formed triangles. In deference to your stated lack of background, I’ve included a lot of algebraic detail; I hope that it isn’t too messy to wade through.
What you have is a sequence of arithmetic progressions whose differences themselves form an arithmetic progression: in your case the differences are $2,4,6,8,\dots$. Moreover, the first terms of your progressions behave in a regular way: they form a sequence, $\langle 1,4,8,13,19,\dots\rangle$, that isn’t arithmetic, but whose difference sequence is. By difference sequence I mean the sequence of differences between consecutive terms: $\langle 3,4,5,6,\dots\rangle$.
Suppose that you have such a collection of arithmetic progressions:
for each $n$, $\langle a_{n,1},a_{n,2},a_{n,3},\dots\rangle$ is an arithmetic progression with constant difference $d_n$;
$\langle d_1,d_2,d_3,\dots\rangle$ is an arithmetic progression with constant difference $d$; and
the sequence $\langle a_{1,1},a_{2,1},a_{3,1},\dots\rangle$ of first terms is not arithmetic, but its difference sequence $\langle a_{2,1}-a_{1,1},a_{3,1}-a_{2,1},a_{4,1}-a_{3,1},\dots\rangle$ is, with constant difference $e$.
Now you build a triangle just as you did with your specific progressions. We can calculate the row sums as follows.
The terms in the $n$-th row are $a_{n,1},a_{n-1,2},a_{n-2,3},\dots,a_{1,n}$. In other words, they are the terms $a_{k,n+1-k}$ for $k=1,2,\dots,n$. Now $a_{k,n+1-k}$ is the $(n+1-k)$-th term in the $k$-th progression, so $a_{k,n+1-k} = a_{k,1} + ((n+1-k)-1)d_k = a_{k,1}+(n-k)d_k,$ and $d_k$ is the $k$-th difference, so $d_k = d_1+(k-1)d.$ Thus, $\begin{align*}a_{k,n+1-k} &= a_{k,1} + (n-k)(d_1+(k-1)d)\\ &= a_{k,1} + (n-k)d_1 + (n-k)(k-1)d\\&= a_{k,1} + nd_1 - kd_1 +(nk - n - k^2 + k)d\\ &= a_{k,1} + n(d_1-d) + ((n+1)d-d_1)k-dk^2, \end{align*}$
and the sum of the terms in the $n$-th row is $\begin{align*} \sum\limits_{k=1}^n a_{k,n+1-k} &= \sum\limits_{k=1}^n \left(a_{k,1} + n(d_1-d) + ((n+1)d-d_1)k-dk^2\right)\\ &= \sum\limits_{k=1}^n a_{k,1} + \sum\limits_{k=1}^n n(d-d_1) + \sum\limits_{k=1}^n ((n+1)d-d_1)k - \sum\limits_{k=1}^n dk^2\\ &= \sum\limits_{k=1}^n a_{k,1} + n^2(d-d_1) + ((n+1)d-d_1)\sum\limits_{k=1}^n k - d\sum\limits_{k=1}^n k^2. \end{align*}$
The summations $\sum\limits_{k=1}^n k = \dfrac12n(n+1)$ and $\sum\limits_{k=1}^n k^2 = \dfrac16n(n+1)(2n+1)$ are well-known, so $\begin{align*} \sum\limits_{k=1}^n a_{k,n+1-k} &= \sum\limits_{k=1}^n a_{k,1} + n^2(d-d_1) + \frac12((n+1)d-d_1)n(n+1) - \frac16dn(n+1)(2n+1)\\ &= \sum\limits_{k=1}^n a_{k,1} + d\left(n^2 + \frac12n(n+1)^2 - \frac16n(n+1)(2n+1)\right) - d_1\left(n^2 + \frac12n(n+1)\right)\\ &= \sum\limits_{k=1}^n a_{k,1} + \frac{d}6\left(6n^2 + 3n(n+1)^2 - n(n+1)(2n+1)\right) - \frac{d_1}2(3n^2+n)\\ &= \sum\limits_{k=1}^n a_{k,1} + \frac{d}6\left(6n^2 + (3n^3+6n^2+3n) - (2n^3+3n^2+n)\right)-\frac{d_1}2(3n^2+n)\\ &= \sum\limits_{k=1}^n a_{k,1} + \frac{d}6(n^3+9n^2+2n)-\frac{d_1}2(3n^2+n)\\ &= \sum\limits_{k=1}^n a_{k,1} + \frac{d}6 n^3 + \frac{3d-3d_1}{2}n^2 + \frac{2d-3d_1}{6}n. \end{align*}$
Now we need to evaluate $\sum\limits_{k=1}^n a_{k,1}$ in terms of $n$. This takes a little work, since the sequence of $a_{k,1}$’s isn’t arithmetic. Let $e_k = a_{k+1,1}-a_k$; we assumed that the sequence $\langle e_1, e_2, e_3,\dots\rangle$ is arithmetic with constant difference $e$, so $e_k = e_1 + (k-1)e$. Thus, $\begin{align*} a_{k,1} &= a_{1,1} + e_1 + e_2 + \dots + e_{k-1}\\ &= a_{1,1} + \sum\limits_{i=1}^{k-1} (e_1 + (i-1)e)\\ &= a_{1,1} + \sum\limits_{i=1}^{k-1} e_1 + e\sum\limits_{i=1}^{k-1} (i-1)\\ &= a_{1,1} + (k-1)e_1 + e\sum\limits_{i=0}^{k-2} i\\ &= a_{1,1} + (k-1)e_1 + \frac12 e(k-2)(k-1)\\ &= a_{1,1} + e_1k - e_1 + \frac{e}2 k^2 - \frac{3e}{2}k+e\\ &= \frac{e}2 k^2 + \frac{2e_1-3e}{2}k + (a_{1,1}+e-e_1), \end{align*}$
and therefore $\begin{align*} \sum\limits_{k=1}^n a_{k,1} &= \sum\limits_{k=1}^n \left(\frac{e}2 k^2 + \frac{2e_1-3e}{2}k + (a_{1,1}+e-e_1)\right)\\ &= \frac{e}{2}\sum\limits_{k=1}^n k^2 + \frac{2e_1-3e}{2}\sum\limits_{k=1}^n k + \sum\limits_{k=1}^n (a_{1,1}+e-e_1)\\ &= \frac{e}{12}n(n+1)(2n+1) + \frac{2e_1-3e}{4}n(n+1) + (a_{1,1}+e-e_1)n\\ &= \frac{e}{12}(2n^3+3n^2+n) + \frac{2e_1-3e}{4}(n^2+n) + (a_{1,1}+e-e_1)n\\ &= \frac{e}6 n^3 + \frac{e_1-e}{2}n^2 + \frac{6a_{1,1}+2e-3e_1}{6}n. \end{align*}$
Finally, then, the $n$-th row sum is $\begin{align*} &\sum\limits_{k=1}^n a_{k,1} + \frac{d}6 n^3 + \frac{3d-3d_1}{2}n^2 + \frac{2d-3d_1}{6}n =\\ &\frac{e}6 n^3 + \frac{e_1-e}{2}n^2 + \frac{6a_{1,1}+2e-3e_1}{6}n + \frac{d}6 n^3 + \frac{3d-3d_1}{2}n^2 + \frac{2d-3d_1}{6}n =\\ &\frac{e+d}6 n^3 + \frac{e_1-e+3d-3d_1}2 n^2 + \frac{6a_{1,1}+2(e+d)-3(e_1+d_1)}6 n,\tag{1} \end{align*}$
which is indeed cubic in $n$.
As a quick check on all this algebra, note that in your original example $a_{1,1}=1$, $d_1=2$, $d=2$, $e_1=3$, and $e=1$, so $(1)$ reduces to $\frac12 n^3 + n^2 - \frac12 n,$ as we know it should.