One alternative argument would be as follows:

1. If there exists a $\rho\ge0$ such that $\Phi\otimes \mathrm{Id}_k(\rho)\not\ge0$, then there will also exist a pure $\vert\chi\rangle$ such that $\Phi\otimes \mathrm{Id}_k(\vert\chi\rangle\langle\chi\vert)\not\ge0$. (This follows immediately from convexity, e.g. by taking an ensemble decomposition of said $\rho$ -- it will contain one such $\vert\chi\rangle$.)

2. Take the Schmidt decomposition of $\vert\chi\rangle$, and denote its Schmidt rank by $\ell$. Clearly, $\ell\le n$. Then, when considering $\Phi\otimes \mathrm{Id}_k(\vert\chi\rangle\langle\chi\vert)$, the extending space (the one with the $\mathrm{Id}$) can be compressed to a space with dimension $\ell$ (spanned by the Schmidt vectors). Call the compressed state $\vert\chi'\rangle$.

3. Clearly, in the compressed space $\Phi\otimes Id_\ell(\vert\chi'\rangle\langle\chi'\vert)\not\ge0$.

One alternative argument would be as follows:

1. If there exists a $\rho\ge0$ such that $\Phi\otimes \mathrm{Id}_k(\rho)\not\ge0$, then there will also exist a pure $\vert\chi\rangle$ such that $\Phi\otimes \mathrm{Id}_k(\vert\chi\rangle\langle\chi\vert)\not\ge0$. (This follows immediately from convexity, e.g. by taking an ensemble decomposition of said $\rho$ -- it will contain one such $\vert\chi\rangle$.)

2. Take the Schmidt decomposition of $\vert\chi\rangle$, and denote its Schmidt rank by $\ell$. Clearly, $\ell\le n$. Then, when considering $\Phi\otimes \mathrm{Id}_k(\vert\chi\rangle\langle\chi\vert)$, the extending space (the one with the $\mathrm{Id}$) can be compressed to a space with dimension $\ell$ (spanned by the Schmidt vectors). Call the compressed state $\vert\chi'\rangle$.

3. Clearly, in the compressed space $\Phi\otimes Id_\ell(\vert\chi'\rangle\langle\chi'\vert)\not\ge0$.

This shows that $n$-positive implies $k$-positive for $k>n$. If, in addition, you want to make sure that $n$-positivity when applied only to the maximally entangled state $\vert\Omega\rangle$ is sufficient, then you can do the following:

4. Assume wlog $\ell = n$. (Otherwise, embed into an $n$-dimensional space.) Write $\vert\chi'\rangle = (\mathrm{I}\otimes M)\vert\Omega\rangle$. Then, $$ (\mathrm{I}\otimes M)\,\big[(\Phi\otimes \mathrm{Id}_n)(\vert\Omega\rangle\langle\Omega\vert)\big]\,(\mathrm{I}\otimes M^\dagger) = (\Phi\otimes \mathrm{Id}_n)(\vert\chi'\rangle\langle\chi'\vert) \not\ge 0\ , $$ which proves that $(\Phi\otimes \mathrm{Id}_n)(\vert\Omega\rangle\langle\Omega\vert)\ge0$ implies $n$-positivity.

(Fun fact on the side: This argument can also be used to show that in order to check CP, it is sufficient to evaluate the action of the channel on any $\vert\chi'\rangle$ with maximal Schmidt rank, since in that case, $M$ is invertible, and the argument works both ways.)

One alternative argument would be as follows:

1. If there exists a $\rho\ge0$ such that $\Phi\otimes \mathrm{Id}_k(\rho)\not\ge0$, then there will also exist a pure $\sigma=\vert\chi\rangle\langle\chi\vert$ such that $\Phi\otimes \mathrm{Id}_k(\sigma)\not\ge0$. (This follows immediately from convexity, e.g. by taking an ensemble decomposition of said $\rho$ -- it will contain one such $\vert\chi\rangle$)

2. Take the Schmidt decomposition of $\vert\chi\rangle$, and denote its Schmidt rank by $\ell$. Clearly, $\ell\le n$. Then the extending space (the one with the $\mathrm{Id}$) can be compressed to a space with dimension $\ell$. Call the compressed state $\vert\chi'\rangle$.

3. Clearly, in the compressed space $\Phi\otimes Id_\ell(\vert\chi'\rangle\langle\chi'\vert)\not\ge0$.

One alternative argument would be as follows:

1. If there exists a $\rho\ge0$ such that $\Phi\otimes \mathrm{Id}_k(\rho)\not\ge0$, then there will also exist a pure $\vert\chi\rangle$ such that $\Phi\otimes \mathrm{Id}_k(\vert\chi\rangle\langle\chi\vert)\not\ge0$. (This follows immediately from convexity, e.g. by taking an ensemble decomposition of said $\rho$ -- it will contain one such $\vert\chi\rangle$.)

2. Take the Schmidt decomposition of $\vert\chi\rangle$, and denote its Schmidt rank by $\ell$. Clearly, $\ell\le n$. Then, when considering $\Phi\otimes \mathrm{Id}_k(\vert\chi\rangle\langle\chi\vert)$, the extending space (the one with the $\mathrm{Id}$) can be compressed to a space with dimension $\ell$ (spanned by the Schmidt vectors). Call the compressed state $\vert\chi'\rangle$.

3. Clearly, in the compressed space $\Phi\otimes Id_\ell(\vert\chi'\rangle\langle\chi'\vert)\not\ge0$.