There were lots of questions about p-norms (or which eventually lead to discussion of only p-norms), in the case of $\mathbb{R}^2$ for $p \geq 1$ it is $\|(x, y)\|_p = (|x|^p+|y|^p)^{1/p}$. For these norms it is known that the highest value of $\pi_{\|\cdot\|_p}$ is $4$ attained at $p=1$ and the lowest value is our $\pi_{\|\cdot \|_p}=\pi=3.14159265\dots$ attained at $p=2$ (discussed in the question [1]).
My question is: what happens in the case of arbitrary norms? In [2] it is said that any centrally symmetric convex set $B$ centred at 0 gives rise to a norm which has $B$ as its unit ball, $\|x\|_B = \inf \{ \lambda^{-1}: \lambda x \in B\}$. As far as I understand, in order to define circumference, $B$ has to have a connected boundary so that there exists a continuous bijective $\gamma: [0,1] \rightarrow \partial B$ which means that $B$ can not be something like $\mathbb{R} \times [-1, 1]$, it has to be a bounded set in terms of $\|\cdot\|_2$. Also I assume that all norms are uniquely defined by their unit ball.
If we take a line segment $ [-1, 1] \times \{0\} $ as our unit ball we get $\pi_{\|\cdot\|} = 2$ since the diameter is $2 = \|(1, 0) - (-1, 0)\|$ and the circumference is $4 = \|(1, 0) - (-1, 0)\| + \|(-1, 0) - (1, 0)\|$. As I mentioned above, $\pi_{\|\cdot\|_1} = 4$. Is it possible to come up with a unit ball $B$ inducing a norm $\|\cdot\|_B$ for which circumference is defined and such that it has $\pi_{\|\cdot\|_B} < 2$ or $\pi_{\|\cdot\|_B} > 4$?
[1] $\pi$ in arbitrary metric spaces
[2] https://www.researchgate.net/publication/1899087_The_Fermat-Torricelli_Problem_in_Normed_Planes_and_Spaces
