I assume $n$ is the number of vertices of the polygon. Then, for any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices for which the set of numbers $A$ is unbounded.
The question is more subtle if you ask for which $n$ it is possible to find a non-constant initial choice of labels for which the set of numbers $A$ is bounded. A partial answer is then: yes is $n$ is a multiple of $4$ (I'm not sure if this is what you wanted, though).
The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$ (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ simple eigenvalues, $$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$
In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $, implying that $L^k v$ is unbounded. Note that $v$ is certainly not a constant vector, as required.
You may ask, for what $n$ it is possible to find a non-constant integer initial labeling $v$ of the vertices for which $A$ is bounded. This is certainly the case if $n$ is a multiple of $4$: there is the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$.
A more subtle question: is the above possible, for a given number $n$ (possibly not a multiple of $4$) ? This is equivalent to the question: are there non-constant integer vectors in the eigenspace of $L$ corresponding to eigenvalues not larger than $1$ in absolute value? The eigenvectors of $L$ are: $$v_k:=(1,\omega^k, \omega^{2k},\dots,\omega^{(n-1)k})^T,\qquad k=1,\dots, n\ ,$$
where $\omega:= \exp(2\pi i /n)$, so the question boils down to asking if a linear combination of the $v_k$ with $|k|\le n/4 $ can be a non-constant integer vector.