I'm not sure how you computed the minimal polynomial, but it's not a necessary step -nor its knowledge will give you, in general, the Jordan canonical form.
Instead, once I had the characteristic polynomial, I would procede as follows. Assume it looks like
$ Q(t) = (t-\lambda)^m \dots $
That is: the eigenvalue $\lambda$ has algebraic multiplicity equal to $m$. That is, the Jordan block of $\lambda$ has size $m\times m$.
Then, I would compute the following ranks:
$ \mathrm{rank} (A - \lambda I) > \mathrm{rank} (A - \lambda I)^2 > \mathrm{rank} (A - \lambda I)^3 > \dots $
Or, what is the same, the dimensions of the following subspaces:
$ \mathrm{Ker} (A - \lambda I) \subset \mathrm{Ker} (A - \lambda I)^2 \subset \mathrm{Ker} (A - \lambda I)^3 \subset \dots $
When I got
$ d_\alpha = \mathrm{dim}\ \mathrm{Ker} (A - \lambda )^\alpha = m $
I would stop. All the information you need about the Jordan canonical form is contained in the sequence of numbers
$ d_i = \mathrm{dim}\ \mathrm{Ker} (A - \lambda I )^i \ , \qquad i = 1, \dots, \alpha \ . $
For instance,
$ d_1 = \mathrm{dim}\ \mathrm{Ker} (A - \lambda I ) = n - \mathrm{rank} (A - \lambda I) $
is the number of "boxes" inside the Jordan block of the eigenvalue $\lambda$. (Here, $n$ is the size of your matrix.)
In your example,
$ d_1 = \mathrm{dim}\ \mathrm{Ker} (A - I ) = 4 - \mathrm{rank} (A - I) = 4- 3 = 1 $
means there is just one box in the Jordan block of $1$. So, it must be $3 \times 3$. That is, it must be
$ \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} $