Claim: The iterated sum $f_k(t_1,\ldots,t_k)$ counts the number of elements the interval $[\emptyset,\lambda]$ of Young's lattice, where $\lambda = (\lambda_1,\lambda_2,\ldots,\lambda_k)$ is the partition determined by $\lambda_{k-i+1} = t_1 + \cdots + t_i$. Equivalently, the function $f_k$ counts the number of subdiagrams of $\lambda$.

For an arbitrary partition $\lambda$, we have
$$|[\emptyset,\lambda]| = \text{det} \left[\binom{\lambda_i + 1}{i-j+1}\right]_{1 \leq i,j \leq k}$$
which is a result due to P. A. MacMahon. The answer to Exercise 149 in Chapter 3 of Stanley's Enumerative Combinatorics, volume 1, 2nd edition provides a good reference of references for this result, with various extensions and specializations, including some of the results mentioned in the comments. For a short visual proof using Lindström-Gessel-Viennot, see Ciucu -
A short conceptual proof of Narayana's path-counting formula.

If the claim is true, MacMahon's result implies
$$\sum_{j_1=0}^{t_1}\sum_{j_2=0}^{t_2+j_1}\cdots\sum_{j_k=0}^{t_k+j_{k-1}} = \text{det} \left[\binom{t_1 + \cdots + t_{k - i + 1} + 1}{i-j+1}\right]_{1 \leq i,j \leq k}$$
which implies $f_k(t_1,\ldots,t_k)$ is a polynomial in $t_1,\ldots,t_k$.

Note that $f_k(t_1,\ldots,t_k)$ counts the number of $(j_1,\ldots,j_k)$ such that $0 \leq j_1 \leq t_1$ and $0 \leq j_{i+1} \leq j_i + t_{i+1}$ for $i \geq 1$. To establish the claim, it suffices to find a bijection between the set of $\mu \subseteq \lambda$ and the set of tuples satisfying the above constraints.

Sketch: Map $\mu \subseteq \lambda$ to $(j_1,\ldots,j_k)$, where $j_i = \lambda_{k-i+1} - \mu_{k-i+1}$. The visual interpretation is that each $j_i$ measures the distance between the walls of the $i$-th row from the bottom of the Young diagrams (English convention) for $\mu$ and $\lambda$. The $t_i$ specify how many boxes are added to the diagram for $\lambda$ in moving from the $(i-1)$-st row from the bottom to the $i$-th row. The constraints express the fact that in going from bottom to top in the diagram, the distance between walls increases by at most $t_i$. For a more direct definition chase, note that $\lambda_{k-i} - \lambda_{k-i+1} = t_{i+1}$. Since $\mu$ is a partition, we have $\mu_{k-i+1} - \mu_{k-i} \leq 0$. Combining the definitions and inequalities gives $j_{i+1} \leq j_i + t_{i+1}$.

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