Foremost, if $\lambda$ is an eigenvalue of $A$, then $(A-\lambda)$ will be a factor of the minimal polynomial (I hope this isn't too hand-wavey, but it is true, so I don't think it is).
We know that $V$ decomposes into the direct sum of the generalized eigenspaces since your field is algebraically closed. Since we know $A$ only has 1 eigenvalue by the minimal polynomial argument,
$$V = G(A, 2) = \ker(A-2I)^7$$
(Where 7 is $\dim V$)
Since the root 2 has multiplicity 3 in the minimal polynomial, we in fact know that $V = \ker(A-2I)^3$, which means that $\dim \ker(A-2I)^3 = 7$, and we also know that $dim\ker(A-2I)^2 \leq 6$, since if it were 7, the minimal polynomial would instead be $(2-t)^2$.
As we know the exact form of the basis, we can discuss the rest in terms of the Jordan chains. We know that the Jordan basis corresponding to generalized eigenvalues of $V$ must all be corresponding to eigenvalue $2$ due to the decomposition of $V$ into its generalized eigenspaces discussed above.
We know of a length 3 Jordan chain because of the multiplicity in the minimal polynomial, so there is some chain in our basis that is $v_1, (A-2I)v_1, (A-2I)^2v_1$, where $(A-2I)^2v_1$ is an eigenvalue by definition of the Jordan basis.
The other chains in our basis could be at most of length 3, which would minimize the number of vectors strictly in the eigenspace. Due to dimension counting, we couldd only have one chain of length 3, and then another of length 1.
So the other two chains, at their longest, are:
$v_2, (A-2I)v_2, (A-2I)^2v_2, v_3$, where $(A-2I)^2v_2$ and $v_3$ are the eigenvalues by definition of the Jordan basis.
We could also have two chains of length 2, which also results in 3 total eigenvectors, one for each chain, 1 chain of length 2 and 2 chains of length 1 (4 total eigenvectors), or 4 chains of length 1 (5 total eigenvectors), all which produce a dimension of the eigenspace greater than or equal to 3.