Let me write total variation norm $$ \|u\|_{TV} = \max_{z\in Q} \langle z, Du\rangle, $$
where $Q$ is the unit ball in $\mathbb R^2$ and $D$ is the corresponding gradient matrix.
I can smooth TV by $$ \max_{z\in Q} \langle z, Du\rangle - \frac{\mu}{2} \|z\|^2, $$ follwing the paper Yu. Nesterov, Smooth minimization of non-smooth functions, Mathematical Programming (2005).
I know if $g(x)=\alpha f$, the dual function is $g^*(x)= \alpha f^*(x/\alpha)$. My question is how to smooth $g$?
Given $\alpha \ge 0$ (lookup the definition of mixed norms), you have $$\alpha\|u\|_{TV} = \alpha \|Du\|_{2,1} = \|\alpha D u \|_{2,1}.$$ So, to smooth $g := \alpha \|.\|_{TV}$, simply replace the linear operator $D$ by the scaled version $\alpha D$.
In particular, you don't need convex conjugates, etc.