The short answer is that $(\rho^{\otimes N})^{T_B}=(\rho^{T_B})^{\otimes N}$.
More explicitly, if $\rho=\sum_{ii'jj'}\rho_{ii',jj'}|i\rangle\!\langle i'|\otimes |j\rangle\!\langle j'|$, then we can write
$$\rho^{\otimes N}=\sum_{I I' JJ'}\rho_{II',JJ'}\bigotimes_{k=1}^N \Big(|i_k\rangle\!\langle i_k'|\otimes |j_k\rangle\!\langle j_k'|\Big),$$
where $I\equiv(i_1,...,i_N)$ and same for $I',J,J'$.
Alternatively, we can write this state highlighting the bipartite structure still present in $\rho^{\otimes N}$ as
$$\rho^{\otimes N}=
\sum_{I I' JJ'}\rho_{II',JJ'}
\left(\bigotimes_{k=1}^N |i_k\rangle\!\langle i_k'|\right)
\otimes
\left(\bigotimes_{k=1}^N |j_k\rangle\!\langle j_k'|\right).$$
The partial transpose operator then acts on it as
\begin{align}
(\rho^{\otimes N})^{T_B} =
\sum_{I I' JJ'}\rho_{II',JJ'}
\left(\bigotimes_{k=1}^N |i_k\rangle\!\langle i_k'|\right)
\otimes
\left(\bigotimes_{k=1}^N |j_k'\rangle\!\langle j_k|\right).\end{align}
In short, this means that the partial transpose effectively operates by changing $\rho_{II',JJ'}\to \rho_{II',J'J}$.
But the coefficients $\rho_{II',JJ'}$ also have a known structure, given by the tensor product operation:
$$\rho_{II',JJ'}=\prod_{k=1}^N \rho_{i_k i_k',j_k j_k'},$$
and the coefficients on the RHS of this expression are exactly what you would get computing explicitly the matrix elements of $(\rho^{T_B})^{\otimes N}$, thus the result follows:
\begin{align}
(\rho^{T_B})^{\otimes N} &=
\left(\sum_{ii',jj'}\rho_{ii',jj'}\big(|i\rangle\!\langle i'|\otimes|j'\rangle\!\langle j|\big)\right)^{\otimes N}\\
&=
\sum_{II'JJ'}\rho_{II',JJ'}
\bigotimes_{k=1}^N \Big(|i_k\rangle\!\langle i_k'|\otimes |j_k'\rangle\!\langle j_k|\Big) \\
&\simeq \sum_{II'JJ'}\rho_{II',JJ'}
\left(\bigotimes_{k=1}^N |i_k\rangle\!\langle i_k'|\right)
\otimes
\left(\bigotimes_{k=1}^N |j_k'\rangle\!\langle j_k|\right)
=(\rho^{\otimes N})^{T_B}.
\end{align}
