Here is a simple recipe.
Let $\mathrm{rk}(C)$ denote the rank of $C$, and put $B:=A-\lambda I$. Write the sequence $\mathrm{rk}(B^0)$, $\mathrm{rk}(B^1)$, $\mathrm{rk}(B^2)$, ... [Remember: $B^0=I$, $B^1=B$.] You get
$$7,\quad 2, \quad 1,\quad 0,\quad 0,\quad\cdots$$
Take the successive differences: $7-2=5$, $1-0=1$, and so on: $$5,\quad 1, \quad 1,\quad 0,\quad 0,\quad\cdots$$ Do it again:
$$4,\quad 0, \quad 1,\quad 0,\quad 0,\quad\cdots$$
This tells you that you have $4$ Jordan blocks of size $1$, $0$ Jordan blocks of size $2$, and $1$ Jordan block of size $3$, and that's it.
More generally, if you put $f(k):=\mathrm{rk}(B^k)$ and $(\Delta g)(k):=g(k)-g(k+1)$, then the number of Jordan blocks of size $k$ is $(\Delta^2 f)(k)$.
Here is a justification. [I'll use the letter $i$ as a subscript; this is not $\sqrt{-1}$.]
Let $A$ be an $n$ by $n$ complex nilpotent matrix, put $V:=\mathbb C^n$, and view $V$ as a $\mathbb C[X]$-module via the formula $fv:=f(A)v$ for $f\in\mathbb C[X]$ and $v\in\mathbb C^n$. Consider the $\mathbb C[X]$-module
$$V_i:=\mathbb C[X]/(X^{i+1}).$$ [For any $f\in\mathbb C[X]$, $(f)$ denotes the ideal generated by $f$.] The endomorphism $v\mapsto Xv$ of $V_i$ has just one Jordan block $J(i+1,0)$ of size $i+1$.
By the Jordan block theory, we have
$$V\simeq\bigoplus_jV_{i(j)},$$ where the sum is finite. Write this symbolically as $$V\simeq\bigoplus_i\ m_i\ V_i,$$ where $m_i$ is the number of times $V_i$ appears in the previous sum, that is, the number of Jordan blocks of size $i+1$. Set
$$n_j:=\dim\frac{X^jV}{X^{j+1}V}\quad.$$
It suffices to verify the
CLAIM: $m_i=n_i-n_{i+1}$.
To prove the claim, form the polynomial
$$V(Y):=\sum\ n_j\ Y^j.$$ We have
$$V_i(Y)=\frac{Y^{i+1}-1}{Y-1}=1+Y+Y^2+\cdots+Y^i,$$
and we must solve
$$\sum\ m_i\ V_i(Y)=\sum\ n_j\ Y^j$$ for the $m_i$, where the $n_j$ are considered as known quantities (almost all equal to zero). Multiplying through by $Y-1$ we get $$\sum\ m_{i-1}\ Y^i-\sum\ m_i=\sum\ (n_{i-1}-n_i)\ Y^i.$$
This proves the claim.