Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element
$$
X:=\Bigg(\,\underbrace{\begin{smallmatrix}
\lambda_1&1&0&0&0\\
0&\lambda_1&1&0&0\\
0&0&\ddots&\ddots&0\\
0&0&0&\lambda_1&1\\
0&0&0&0&\lambda_1\\
\end{smallmatrix}}_{n_1}\,\Bigg)
\oplus
\Bigg(\,\underbrace{\begin{smallmatrix}
\lambda_2&1&0&0&0\\
0&\lambda_2&1&0&0\\
0&0&\ddots&\ddots&0\\
0&0&0&\lambda_2&1\\
0&0&0&0&\lambda_2\\
\end{smallmatrix}}_{n_2}\,\Bigg)
\oplus
\cdots\oplus
\Bigg(\,\underbrace{\begin{smallmatrix}
\lambda_k&1&0&0&0\\
0&\lambda_k&1&0&0\\
0&0&\ddots&\ddots&0\\
0&0&0&\lambda_k&1\\
0&0&0&0&\lambda_k\\
\end{smallmatrix}}_{n_k}\,\Bigg)
$$
I claim that for generic $\lambda_i$ that element generates your algebra.

First of all, $(X-\lambda_j)^{n_j}$ has zero for its $j$th component.
Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand (say the $i$th summand). The remaining term is Toeplitz and upper triangular. More precisely, it is given by
$$
A:=\left(\begin{matrix}
a&b&c&d&e\\
0&a&b&c&d\\
0&0&\ddots&\ddots&c\\
0&0&0&a&b\\
0&0&0&0&a\\
\end{matrix}\,\right)
$$
with $a=\prod_{j\not =i}(\lambda_i-\lambda_j)^{n_j}$
and $b=$ (some horrible expression which is obviously positive if all the $(\lambda_i-\lambda_j)$ are positive, and therefore non-zero if the $\lambda_i$ are chosen generically).

Then it's a matter of playing around to see that the above matrix and its adjoint generate all of $M_{n_i}(\mathbb C)$.
The first operation is to consider $N:=A^2-aA$, which is now nilpotent with non-zero terms on the subdiagonal. Then you take powers of $N$. Then you take powers of $N^*$, etc.