For any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices of the $n$-gone 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 integer choice of labels for which the set of numbers $A$ is bounded. The answer is then: it is possible if and only if $n$ is either a multiple of $4$ or a multiple of $6$.
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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 $n$ real eigenvalues (repeated according to their multiplicity, which is either $1$ or $2$)
$$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$
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} $ in the spectral decomposition, implying that $A=\{L^k v\}_{k\ge0}$ is unbounded. Note that $v$ is certainly not a constant vector, as required. Also note that for even $n$, $\lambda_{n/2}=-3$ corresponds to the example in Barry Cipra's comment.
For what $n$ it is possible to find non-constant integer labels $v=(v_1,\dots,v_n)$ of the vertices for which $A=\{L^k v\}_{k\ge0}$ is bounded? This is certainly the case if $n$ is a multiple of $4$: then we have the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$.
If $n$ is a multiple of $6$, $L$ is singular with a $2$ dimensional kernel spanned by the $6$-periodic vectors $(1,0,-1,-1,0,1,\dots)$ and $(1,1,-1,-1,1,1,\dots)$. Of course these generate bounded orbits.
Conversely, assume that $n$ is such that there exists a non-constant $v\in \mathbb{Z}^n$
that generates a bounded orbit $\{L^k v\}_k $ . If $L$ is singular, $n$ must be a multiple of $6$, as it follows from the above expression for the eigenvalues. If $L$ is not singular the orbit $\{L^k v\}_k $ is periodic, that is, $L^m v = v$ for some $m$. So $1$ is one of the eigenvalues of $L^m$, which are of the form $\lambda_k^m$, and this is only possible if $\lambda_k=-1$ or $\lambda_k=1$. In the former case, $k=0$ and $v$ is a simple eigenvalue of $L$ with constant coordinates, which is not the case. In the latter case, $\cos(2\pi k/n)=0$, which implies that $n=4k$.
