When answering this question about finding the open unit ball $\mathscr{B} := \{ x \in \mathbb{R}^2: \| x \| < 1\}$ of the "composite" norm $$ \| \cdot \|: \mathbb{R}^2 \to \mathbb{R}, \ (x,y) \mapsto a \| (x,y) \|_1 + \frac{b}{2} \| (x,y) \|_{\infty}. $$ I thought of the following question. In the above question one has $\Omega := \mathbb{R}^2$, $a := \frac{1}{3}$ and $b := \frac{4}{3}$ but those aren't important for my question. All that matters is $a,b > 0$, as verified in this question.
It turns out that $\mathscr{B}$ is a octagon (as intersection of two rotated squares, as they are the geometric interpretations of $\| \cdot \|_1$ and $\| \cdot \|_{\infty}$ (is that really true?), which can be seen in the diagram appended to my answer to the first mentioned question).
My question is if (and how) one can find out which shape (polygon?) $\mathscr{B}$ corresponds for a composite norm of the form $$ \| \cdot \| := \sum_{k = 1}^{\infty} \alpha_k \| \cdot \|_{x_k}, \qquad \text{where } \alpha_k \ge 0, x_k \in [1, \infty]. $$ As @CalvinKohr points out in the comments, we can normalize this representation: $\sum_{k} \alpha_k = 1$ such that the sum is well defined i.e. converges.
This question seems to be related but I don't know how the Minkowski functional would relate to this problem even though it was briefly covered in my Functional Analysis course. It remarks that a polygon with a odd number of vertices can not occur because of the symmetry of the norm. As you can see in the last example below, other shapes than octagons are possible. Can $\mathscr{B}$ be another polygon with an even number of vertices?
Maybe this is related to the concept of polyhedral norms?
One special case Cosider the norm $\mathfrak{p}_n(x,y) := \sum_{k = 1}^{n} \| (x,y) \|_{k}$. If we graph it and intersect it with a plane $z = \ell$ for $\ell > 0$ we obtain the the shape of $\mathscr{B}$. I graphed $\mathfrak{p}_n$ for $n \in \{1, \ldots, 5\}$ and one observes that shapes of $\mathscr{B}$ are 4-gons that "get more convex" and converge to some circle.
This suggests it might by only interesting to at norms whose $\mathscr{B}$ is a polygon i.e. $\mathscr{B}$s with straight lines. Are those just produced by $\| \cdot \|_1$ and $\| \cdot \|_{\infty}$?.
At OP's request, I expand my comments into an answer.
Even though it doesn't have the form stated in the question, a particularly simple example of a norm whose unit circle is an $m$-sided regular polygon for even $m>2$ is
$$\| (x,y) \|_{1,m} = \frac{1}{2\cot\left(\frac{\pi}{m}\right)}\sum_{n=1}^m\left|x\sin\left(\frac{2\pi n}{m}\right)+y\cos\left(\frac{2\pi n}{m}\right)\right|.$$
In particular we have that $\| (x,y) \|_{1,4}=|x|+|y|=\| (x,y) \|_{1}$ is the usual taxicab norm. As was pointed out in the question, a similar norm for odd-sided regular polygons cannot exist since they don't have the required central symmetry.
Proof: As the norm is invariant under rotations of $\frac{2\pi}{m}$ radians, we can restrict ourselves to the set of points whose argument is strictly between $0$ and $\frac{2\pi}{m}$. Within that sector none of the summands become zero, so we can replace $|\cdots|$ by either $(\cdots)$ or $−(\cdots)$, according to whether the argument inside the bars is positive or negative.
Since then the expression turns into an ordinary linear function of $x$ and $y$, it is obvious that the set of points with norm $1$ will be a straight line segment. The result follows by symmetry and the fact that the unit circle must be connected.
Finally, to make the resulting regular polygon have a vertex at $(1,0)$, we divide the sum by the normalization constant $\sum_{n=1}^m\left|1\cdot \sin\left(\frac{2\pi n}{m}\right)+0\cdot \cos\left(\frac{2\pi n}{m}\right)\right| = 2\cot\left(\frac{\pi}{m}\right)$.
This link shows the unit circle corresponding to these norms for even $m$ between $4$ and $12$, together with a circumscribed (ordinary) circle for comparison purposes. The slider can be used to change the value of $m$.