To complement Erwin Kalvelagen's answer, we have the following optimization problem in $\mathrm x \in \mathbb R^n$
$$\begin{array}{ll} \text{minimize} & \|\mathrm A \mathrm x - \mathrm b\|_\infty\end{array}$$
where $\mathrm A \in \mathbb R^{m \times n}$ and $\mathrm b \in \mathbb R^m$ are given. Let $\mathrm a_i^\top \in \mathbb R^n$ denote the $i$-th row of $\rm A$.
Introducing decision variable $t \in \mathbb R$ and rewriting in epigraph form, we then obtain the following constrained optimization problem in $\mathrm x \in \mathbb R^n$ and $t \in \mathbb R$
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & \|\mathrm A \mathrm x - \mathrm b\|_\infty \leq t\end{array}$$
which can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & \displaystyle\max_{1 \leq i \leq m} | \mathrm a_i^\top \mathrm x - b_i | \leq t\end{array}$$
which can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & | \mathrm a_1^\top \mathrm x - b_1 | \leq t\\ & | \mathrm a_2^\top \mathrm x - b_2 | \leq t\\ & \qquad \vdots\\ & |\mathrm a_m^\top \mathrm x - b_m | \leq t\end{array}$$
which can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & -t \leq \mathrm a_1^\top \mathrm x - b_1 \leq t\\ & -t \leq \mathrm a_2^\top \mathrm x - b_2 \leq t\\ & \qquad\quad \vdots\\ & -t \leq \mathrm a_m^\top \mathrm x - b_m \leq t\end{array}$$
which can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & -t 1_m \leq \mathrm A \mathrm x - \mathrm b \leq t 1_m\end{array}$$
where the optimal $\rm x$ and the optimal $t$ are the minimizer and the minimum of the original problem, respectively.
The problem is convex, as demonstrated by the following second-order cone formulation, obtained by introducing $y_i$ to represent $1/x_i$:
\begin{align}
&\text{minimize} &\sum_i c_i y_i \\
&\text{subject to} & A x &\le b \\
&& z &= \sqrt 2 \\
&& 2 x_i y_i &\ge z^2 &&\text{for all $i$} \tag1\label1\\
&& x_i &\ge 0 &&\text{for all $i$}\\
&& y_i &\ge 0 &&\text{for all $i$}
\end{align}
Constraint \eqref{1} is a rotated second-order cone constraint. Everything else is linear.
Best Answer
Hint:
The problem is equilvalent to: $$\min_x \sum_{i=1}^n \max(x_i, -x_i)+\max(x_1, \ldots, x_n)$$
subject to $$Ax=b$$
How would you rewrite sum of minimax as a linear programming problem.
Remark: $Ax=b$ doesn't affect the main problem of interest. Just copy it down as it is.
Edit:
$$\min \sum_{i=1}^n z_i + t$$
subject to
$$z_i \ge x_i , \forall i \in \{1, \ldots, n\}$$ $$z_i \ge -x_i , \forall i \in \{1, \ldots, n\}$$ $$t \ge x_i , \forall i \in \{1, \ldots, n\}$$ $$t \ge -x_i , \forall i \in \{1, \ldots, n\}$$ $$Ax=b$$
You can also write them in terms of vector form as you have shown in the comment as well.