Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.
There is an analogue of the Zeckendorf representation: if you have the sequence of $n$-bonacci numbers, then each positive integer $k$ has a unique representation into a sum of some of those, with no $n$ consecutive terms.
Each expansion of $k$ as a sum of distinct $n$-bonacci numbers can be obbtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ ones. It can even be done in a way that all representations obtained in the process contain only distinct numbers.
For an odd $n$ (in particular, for $n=3$), all such representations are counted in the product under consideration with the same sign, so it counts just the number of such representations which is indeed unbounded.
FFor even $n$, one can show that this process is similar to what happens with ordinary Fibonacci (to each $n$-Zeckendorf representation, one can put into correspondence a Zeckendorf representation of a different number to which just the same set of operations is applicable), so the resulting coefficients are still bounded by $1$. (I may fill the details later, if needed.)
Addendum. However, it seems that one can find products of such kind which will have bounded coefficients! E.g., $$ \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}). $$
