According to the request of the host, I propose here an answer without using the notion of conjugate function. However I have to admit that the idea behind the computations is of conjugate function.
Recall that the Lagrangian function is
\begin{align}
L(x,z,\lambda)=\sum_{i=1}^{3} \langle z_i,\lambda_i\rangle +||z_i||-\langle x, \sum_{i=1}^3\lambda_i\rangle +\sum_{i=1}^3 \langle \lambda_i, a_i\rangle,
\end{align}
where $x, z_i, \lambda_i \in \mathbb R^n$ for $i=1,2,3$.
We claim that the dual problem is
$$\sup_{\lambda} \inf_{x,z} L(x,z,\lambda)
=\sup_{\lambda\in D} \sum_{i=1}^3 \langle \lambda_i, a_i\rangle. \label{eq1} \tag{1}$$
where
\begin{align}
D=\{\lambda=(\lambda_1, \lambda_2, \lambda_3): \sum_{i=1}^3\lambda_i=0 \text{ and }
||\lambda_i||_*\leq 1 \text{ for } i=1,2,3\}.
\end{align}
Here $||\cdot||_*$ is the dual norm of $||\cdot||$ in $\mathbb R^n$ and defined as
\begin{align}
||u||_*:=\sup_{||v||\leq 1} \langle u,v\rangle.
\end{align}
Proof of \eqref{eq1}.
We now focus on $\inf_{x,z} L(x,z, \lambda)$.
We claim that
$$
\inf_x -\langle x, \sum_{i=1}^3\lambda_i\rangle =
\begin{cases}
0, & \text{ if } \sum_{i=1}^3\lambda_i=0,\\
-\infty, & \text{ otherwise.}
\end{cases}
\label{eq2} \tag{2}$$
and
$$
\inf_{z_i} \langle z_i, \lambda_i \rangle +||z_i||=
\begin{cases}
0, & \text{ if } ||\lambda_i||_* \leq 1,\\
-\infty, & \text{ if } ||\lambda_i||_* > 1.
\end{cases}
\label{eq3} \tag{3}
$$
If \eqref{eq2} and \eqref{eq3} hold true, we obtain the dual problem \eqref{eq1}.
Since the host already know how to prove \eqref{eq2}, we now only concentrate on \eqref{eq3}.
To simplify the notations, in \eqref{eq3} we shall remove the index $i$, i.e.
we are going to prove
$$
\inf_{z} \langle z, \lambda\rangle +||z||=
\begin{cases}
0, & \text{ if } ||\lambda||_* \leq 1,\\
-\infty, & \text{ if } ||\lambda||_* > 1.
\end{cases}
\textbf{ for $z, \lambda$ are in $\mathbb R^n$ from here on.}
\label{eq4} \tag{4}
$$
Let $\lambda'=-\lambda$, we have
$$\inf_{z} \langle z, \lambda\rangle +||z||=-\sup_{z} \langle z, \lambda'\rangle -||z||.$$
Case 1. If $||\lambda'||_*\leq 1$, we have
$$\sup_{z} \langle \frac{z}{||z||}, \lambda'\rangle \leq ||\lambda'||_*\leq 1.$$
Here the first inequality holds due to the definition of dual norm.
Thus
$$\sup_{z} \langle z, \lambda'\rangle -||z||\leq 0.$$
Furthermore, the equality holds if we take $z=0$.
Case 2. If $||\lambda'||_*> 1$, from the definition of dual norm,
we deduce the existence of some $z$ such that $||z||\leq 1$
and $\langle z,\lambda'\rangle >1$.
Now, we have
$$\sup_{t\in \mathbb R} \langle tz, \lambda'\rangle-||tz||
=\sup_{t\in \mathbb R} t( \langle z, \lambda'\rangle-||z||) =+\infty.$$
Note that $||\lambda'||_*=||\lambda||_*$, all together tell us that \eqref{eq4} is true.
Hence, \eqref{eq1} is true.
