The task is to minimize $|x1 - b|_{l1}$ in $l_1$ norm. Here $1 \in R^{n}$ is a vector of ones, $x \in R$ is a scalar, $b \in R^{n}$.
First of all, I wrote it in a general form without specifying the norm: minimize $|x1-b|$.
Then by the introduction of the equality constraint, it can be written in an equivalent form as: $$\text{minimize }|y|$$ $$\text{subject to: } y=x1-b $$.
The dual of the problem is $L(\lambda) = |y|+\lambda^T(y-x1+b)$. Taking the infimum we get $g(\lambda) = \inf_{x,y} L(\lambda) = \inf_{x,y}|y|+\lambda^T(y-x1+b) = \inf_{x}\lambda^T(b-x1) \text{ subject to }|\lambda|_{*} \le 1$.
So, the dual problem is $$\text{maximize }\inf_{x}\lambda^T(b-x1)$$ $$\text{subject to: } |\lambda|_{*} \le 1 $$.
Where $|.|_{*}$ is the dual norm.
Is this right so far? What should I do next?
The little issue with $y$ in the Lagrangian aside, the steps so far are correct. The next step is to focus on $\inf_x \lambda^T(x1) = \inf_x \lambda^T1 x$. When is the infimum finite?
Update based on the discussion: $$\inf_x \lambda^T1 x = \begin{cases} 0 & \text{if } \lambda^T1 = 0 \\ -\infty & \text{otherwise.}\end{cases}$$