Theorem:
Let $\X$ and $\Y$ be complex Euclidean spaces with $\dim(\X) = n$ and let $\Phi\in\Trans(\X,\Y)$. Let $\norm{\cdot}_1$ denote the usual trace norm of an density operator, and $\triplenorm{\cdot}_1$ the operator norm of a channel.
\[
\triplenorm{\Phi}_1 \leq \norm{J(\Phi)}_1 \leq n
\triplenorm{\Phi}_1. \]
Proof:
Since the Choi representation of $\Phi$ is given by $J(\Phi)=(\Phi\otimes\I_{\Lin(\X)})(\vec(\I_\X)\vec(\I_\X)^\ast)$, then the trace norm is given by and also satisfies
\[\begin{align*}
\norm{J(\Phi)}_1&=\norm{(\Phi\otimes\I_{\Lin(\X)})(\vec(\I_\X)\vec(\I_\X)^\ast)}_1 \\
&\leq \norm{(\Phi\otimes\I_{\Lin(\X)})}_1\norm{\vec(\I_\X)\vec(\I_\X)^\ast}_1 \\
&=\norm{(\Phi\otimes\I_{\Lin(\X)})}_1n \\
&=n \triplenorm{\Phi}_1,
\end{align*}\]
where the last two lines from $\norm{\vec(\I_\X)\vec(\I_\X)^\ast}_1=n$ and the definition of the completely bounded trace norm $\triplenorm{\Phi}_1 :=\norm{(\Phi\otimes\I_{\Lin(\X)})}_1$.
Now consider an alternate characterization of the completely bounded trace norm:
\[
\triplenorm{\Phi}_1=max\{\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_1 : \rho_0,\rho_1\in\Density(\X)\}.
\]
Since this norm satisfies the property
\[
\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_\infty\leq\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_0}}_\infty\norm{J(\Phi)}_1\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_1}}_\infty,
\]
and the spectral norm $\norm{A}_\infty$ of an operator $A$ is given by the largest singular value of $A$, then it follows that $\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_a}}_\infty\leq1$ with equality holding in the case where $\rho_a$ is a pure state. Therefore,
\[
\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_\infty\leq\norm{J(\Phi)}_1,
\]
implying that
\[
\triplenorm{\Phi}_1=max\{\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_1 : \rho_0,\rho_1\in\Density(\X)\}\leq \norm{J(\Phi)}_1.
\]
Putting these two bounds together then gives
\[
\triplenorm{\Phi}_1 \leq \norm{J(\Phi)}_1 \leq n
\triplenorm{\Phi}_1.
\]
Let $\X$ and $\Y$ be complex Euclidean spaces with $\dim(\X) = n$ and let $\Phi\in\Trans(\X,\Y)$. Let $\norm{\cdot}_1$ denote the usual trace norm of an density operator, and $\triplenorm{\cdot}_1$ the operator norm of a channel.
\[
\triplenorm{\Phi}_1 \leq \norm{J(\Phi)}_1 \leq n
\triplenorm{\Phi}_1. \]
Proof:
Since the Choi representation of $\Phi$ is given by $J(\Phi)=(\Phi\otimes\I_{\Lin(\X)})(\vec(\I_\X)\vec(\I_\X)^\ast)$, then the trace norm is given by and also satisfies
\[\begin{align*}
\norm{J(\Phi)}_1&=\norm{(\Phi\otimes\I_{\Lin(\X)})(\vec(\I_\X)\vec(\I_\X)^\ast)}_1 \\
&\leq \norm{(\Phi\otimes\I_{\Lin(\X)})}_1\norm{\vec(\I_\X)\vec(\I_\X)^\ast}_1 \\
&=\norm{(\Phi\otimes\I_{\Lin(\X)})}_1n \\
&=n \triplenorm{\Phi}_1,
\end{align*}\]
where the last two lines from $\norm{\vec(\I_\X)\vec(\I_\X)^\ast}_1=n$ and the definition of the completely bounded trace norm $\triplenorm{\Phi}_1 :=\norm{(\Phi\otimes\I_{\Lin(\X)})}_1$.
Now consider an alternate characterization of the completely bounded trace norm:
\[
\triplenorm{\Phi}_1=max\{\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_1 : \rho_0,\rho_1\in\Density(\X)\}.
\]
Since this norm satisfies the property
\[
\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_\infty\leq\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_0}}_\infty\norm{J(\Phi)}_1\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_1}}_\infty,
\]
and the spectral norm $\norm{A}_\infty$ of an operator $A$ is given by the largest singular value of $A$, then it follows that $\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_a}}_\infty\leq1$ with equality holding in the case where $\rho_a$ is a pure state. Therefore,
\[
\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_\infty\leq\norm{J(\Phi)}_1,
\]
implying that
\[
\triplenorm{\Phi}_1=max\{\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_1 : \rho_0,\rho_1\in\Density(\X)\}\leq \norm{J(\Phi)}_1.
\]
Putting these two bounds together then gives
\[
\triplenorm{\Phi}_1 \leq \norm{J(\Phi)}_1 \leq n
\triplenorm{\Phi}_1.
\]