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 $$ \left(\begin{matrix} a&b&c&d&e\\ 0&a&b&c&d\\ 0&0&\ddots&\ddots&c\\ 0&0&0&a&b\\ 0&0&0&0&a\\ \end{matrix}\,\right) $$ 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 $M_{n_i}(\mathbb C)$.