Example of norm on $\mathbb{R}^2$ that's NOT absolutely monotonic.

Consider a sheer $A(x,y):=(x,y-x)$ and then the norm $\|(x,y)\|:=\|A(x,y)\|_1$. Then $\|(1,0)\|=2$ but $\|(1,1)\|=1$.

Interestingly though, for every norm $\|\cdot\|:\mathbb{R}^2\to[0,\infty)$ there exists a linear map $A:\mathbb{R}^2\to\mathbb{R}^2$ such that $\|\cdot\|\circ A$ is an absolutely monotonic norm.

To prove this, note that a norm is absolutely monotonic if (and only if, may I add) the smallest axis-aligned rectangle containing the unit disk touches the unit circle at the intersections with the $x$- and $y$-axis. This is a consequence of the triangle inequality. So if we find vectors $v$ and $u$ on the unit sphere such that the unit disk is contained in the parallellogram $[-1,1]v\times[-1,1]u$, then the unique linear map that sends $v$ to $(1,0)$ and $u$ to $(0,1)$ suffices.

In order to find such $v$ and $u$, we will use the intermediate value theorem. For any angle $\theta$ let $v(\theta)$ be the unique vector on the unit sphere with angle $\theta$ with respect to the origin. The uniqueness follows from the scaling property and the non-degeneracy of norms, and the continuity follows from the triangle inequality. Then let $u(\theta)$ be the vector on the unit sphere furthest to the left of the line generated by $v(\theta)$. A little bit of a problem here is that $u(\theta)$ is not always uniquely defined, and $\theta\mapsto u(\theta)$ is only continuous at points where it is uniquely defined. However, for now just pretend that it is a well-defined continuous function, and I will come back to this later.

We now find that $\mathbb{R}v\times[-1,1]u$ contains the unit disk. Most importantly the intersection of the unit disk with $(1,\infty)v\times[-1,1]u$ is fully contained in either $(1,\infty)v\times[-1,0]u$ or $(1,\infty)v\times[0,1]u$, due to the triangle inequality. If for some $\theta$ this intersection lies in one of the two, then if we move $\theta$ such that the new $v(\theta)$ is where the old $u(\theta)$ was, the intersection will lie in the other one. For some $\theta$ in between, we must therefore have that the intersection lies in neither. Therefore the unit disk will be contained in $[-1,1]v\times[-1,1]u$.

Finally, to come back to the little problem. For such problematic $\theta$ you can imagine fixing $v(\theta)$ while moving $u(\theta)$ continuously along the set of all possible values.

This is a little bit heuristic, but I hope it is convincing anyways.

I would like to give an answer similar to Smiley-Craft but from a different point of view:

The standard norm on $\mathbb{R}^2$ is given by $\|(a,b)\|= \sqrt {a^2+b^2}$. We can define a similar norm using a different basis.

Given $u,v$ a basis of $\mathbb{R}^2$. We can define the norm of $w$ by $$\|w\| = \sqrt{a^2+b^2}$$ where $w=au+bv$.

It is an easy exercise to show that this is indeed a norm (same proof as in the standard norm case).

Since we can change coordinates it will be really easy to define a non-monotonic norm:

For instance choose $u=(\frac{1}{2},0)$ and $v=(1,1)$ then $\|(1,0)\| = 2$ while $\|(1,1)\| = 1$.

Side note: It is an interesting question whether every norm is absolutely monotonic with respect to some basis. I would guess the answer is yes but I don't know how to approach this question yet.