Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.

There is an [analogue of the Zeckendorf representation][1]: 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 obtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ lots of $n$-bonacci numbers. It can even be done in a way that all representations obtained in the process contain only distinct numbers (in line with the product in current MO question). 

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. To this end, consider the modified product (where $y$ registers number of products contributing to $x^k$)
\begin{align*}
\prod_{i\geq3}(1+y\,x^{T_i})
&=1+yx+yx^2+\cdots+(y^5+y^3+y)x^{81}+(y^6+y^4+y^2)x^{82}+\cdots
\end{align*}
in which either the powers of $y$ are all of the same parity (e.g. coefficients of $x^{81}$ or $x^{82}$).

For 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.) By contrast, take the case $n=4$ and let $J_k$ be given by $\sum_{k\geq0}J_kx^k=\frac{x^3}{1-x-x^2-x^3-x^4}$. The corresponding product
\begin{align*}
\prod_{i\geq4}(1+y\,x^{J_i})
&=1+yx+yx^2+\cdots+(y^7+y^4+y)x^{208}+(y^5+y^2)x^{209}+\cdots
\end{align*}
shows that the powers of $y$ can take up any parity. 


**Addendum.** However, it seems that one can find products of such kind which will have bounded coefficients! E.g., 
$$
  \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}).
$$

[1]: https://www.fq.math.ca/Papers1/44-4/quarttamas04_2006.pdf