**The answer is that that maximal possible value of $c$ is $-3$.**

In fact, this value is obtained if we start with the numbers $1,2,3,4,5$, and following the process:
$$1, 2, 3, 4, 5 \xrightarrow{\alpha} 0, 1, 4, 5 \xrightarrow{\alpha} 0, 1, 2 \xrightarrow{\alpha} -1, 0 \xrightarrow{\beta} -3. \quad (*)$$
We will prove now that no matter which are initially the numbers, we cannot have a number smaller than $-3$ during the process.

To this end, we will find an invariant under the movements. Let $w$ be the real root of the polynomial
$$P x = x^3 + x^2 - 1.$$ Then,
$$w^3 + w^2 = 1 \Rightarrow w^{n+1} + w^n = w^{n-2}. \quad (1)$$
Moreover, the polynomial $P x$ divides $Q x = x^5 + x - 1$, so
$$Q w = 0 \Leftrightarrow w^5 + w = 1 \Leftrightarrow w^{k+4} + w^k = w^{k-1} \quad (2).$$
From (1) and (2) we observe that if we consider one extra board which will have the numbers of our board raised to the $w$ power, then the sum of the numbers in the new board is invariant under $\alpha$) and $\beta$). For example, for the sequence in (*) the extra board will have the numbers,
$$w^1, w^2, w^3, w^4, w^5 \xrightarrow{\alpha} w^0, w^1, w^4, w^5 \xrightarrow{\alpha} w^0, w^1, w^2 \xrightarrow{\alpha} w^{-1}, w^0 \xrightarrow{\beta} w^{-3}.$$
We observe, for example, that after the first movement the sum remains the same since $w^3 + w^2 = w^0$.

It is then enough to start with the minimal sum in the extra board, which is of course bigger than
$$w^1 + w^2 + w^3 + w^4 + \dots = \frac{w}{1-w} = \frac{w}{w^5} = w^{-4}.$$
So, in the initial board all the numbers are bigger than $-4$, so $c \ge -3$.