If $d$ is the maximal exponent of a factor $(x-\lambda)^d$ of $m_A$, it means that there is at least one Jordan block associated with $\lambda$ of size $d$, and any Jordan block associated with $\lambda$ has size no greater than $d$.

We know that $n\ge 8$, or else $A$ can't have a degree $8$ minimal polynomial. The only freedom left for constructing the remaining Jordan blocks are:

1. the number of basis vectors associated with 1 (this also determines the number of basis vectors associated with 2)
2. the sizes of blocks associated with 1
3. the sizes of blocks associated with 2

Consider the case where there are $s$ basis vectors associated with $\lambda$. Let $p(n)$ be the number of partitions of $n$ where each term does not exceed 4. Then $p(s-4)$ is number of possible configurations for blocks associated with $\lambda$ (the first 4 slots must be occupied by a size 4 block, hence the $s-4$). Thus, the total number of configurations is $\sum_{s=4}^{n-4} p(s-4)p(n-s-4)$. Let's call this number $C(n)$.

Having done the calculations, I found that $C(12)=20$ and $C(13)=34$, so the $n$ in your original problem formulation must be $12$ and the answer must be $20$ instead of $25$.

If $d$ is the maximal exponent of a factor $(x-\lambda)^d$ of $m_A$, it means that there is at least one Jordan block associated with $\lambda$ of size $d$, and any Jordan block associated with $\lambda$ has size no greater than $d$.

We know that $n\ge 8$, or else $A$ can't have a degree $8$ minimal polynomial. The only freedom left for constructing the remaining Jordan blocks are:

1. the number of basis vectors associated with 1 (this also determines the number of basis vectors associated with 2)
2. the sizes of blocks associated with 1
3. the sizes of blocks associated with 2

Consider the case where there are $s$ basis vectors associated with $\lambda$. Let $p(m)$ be the number of partitions of $m$ where each term does not exceed 4. Then $p(s-4)$ is the number of possible configurations for blocks associated with $\lambda$ (the first 4 slots must be occupied by a size 4 block, hence the $s-4$). Thus, the total number of configurations is $\sum_{s=4}^{n-4} p(s-4)p(n-s-4)$. Let's call this number $C(n)$.

Having done the calculations, I found that $C(12)=20$ and $C(13)=34$, so the $n$ in your original problem formulation must be $12$ and the answer must be $20$ instead of $25$.