The answer to "Generality" at least seems to be yes. Fix $c \leq n \geq 2$ and let $\mathcal{D}$ be the variety of triples $(U, A, B)$, where $U$ is a codimension $c$ subspace of $M_n(\mathbf{C})$ and $(A, B) \in \mathcal{C}$ satisfies $A \in U$; this is a subvariety of $\operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times M_n(\mathbf{C}) \times M_n(\mathbf{C})$.

Proposition. If $c < n$ then $\mathcal{D}$ is irreducible of dimension $(n^2 - c - 1)c + n^2 + n$, and if $c = n$ then $\mathcal{D}$ has two irreducible components, each of the same dimension $n^3 (= (n^2 - n - 1)n + n^2 + n)$.

This is enough to establish "Generality": indeed, there is a natural surjective map $\pi\colon \mathcal{D} \to \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c)$ given by $\pi(U, A, B) = U$, and for a fixed $U$ the fiber $\pi^{-1}(U)$ is isomorphic to $\mathcal{C}_U$. So by generic flatness, it would follow that for generic $U$ we have $\dim \mathcal{C}_U = \dim \mathcal{D} - \dim \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) = n^2 + n - c$, as desired.

Claim 1. Every irreducible component of $\mathcal{D}$ is of dimension $\geq (n^2 - c - 1)c + n^2 + n$.

To see this, note that there is a Cartesian diagram $$\require{AMScd} \begin{CD} \mathcal{D} @>{}>> \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times \mathcal{C}\\ @VVV @VVV \\ \mathcal{I} @>{i}>> \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times M_n(\mathbf{C}) \end{CD}$$ where $\mathcal{I} = \{(U, A)\colon A \in U\}$ and the right vertical map is given by $(U, A, B) \mapsto (U, A)$. The map $i$ is a closed embedding of codimension $c$ between smooth varieties, so it is locally cut out by $c$ equations. In particular, $\mathcal{D}$ is locally cut out of $\operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times \mathcal{C}$ by $c$ equations, and the claim follows.

Next, define $\alpha\colon \mathcal{D} \to M_n(\mathbf{C})$ by $\alpha(U, A, B) = A$. For any $A \in M_n(\mathbf{C})$, let $Z(A)$ be the space of $B \in M_n(\mathbf{C})$ such that $[A, B] = 0$, and let $z_A = \dim Z(A)$. Note $n \leq z_A \leq n^2$. For each $n \leq z \leq n^2$, let $\Omega_z$ be the locally closed subvariety of $M_n(\mathbf{C})$ consisting of $A$ such that $z_A = z$, and note that $\Omega_n$ is open and dense in $M_n(\mathbf{C})$.

Claim 2. If $z > n$, then $\dim \Omega_z + z < n^2 + n$.

Indeed, let $\mathrm{pr}_1\colon \mathcal{C} \to M_n(\mathbf{C})$ be the first projection, and note that the closure of $\mathrm{pr}_1^{-1}(\Omega_z)$ is a proper subvariety of $\mathcal{C}$, and since $\mathcal{C}$ is irreducible of dimension $n^2 + n$ it follows that $\dim \mathrm{pr}_1^{-1}(\Omega_z) < n^2 + n$. The map $\mathrm{pr}_1^{-1}(\Omega_z) \to \Omega_z$ has fibers of dimension $z$, so the claim follows.

Claim 3. i) $\alpha^{-1}(\Omega_n)$ is irreducible of dimension $(n^2 - c - 1)c + n^2 + n$.

ii) If $n < z < n^2$, then $\dim(\alpha^{-1}(\Omega_z)) < (n^2 - c - 1)c + n^2 + n$.

iii) $\dim(\alpha^{-1}(\Omega_{n^2} - \{0\})) = (n^2 - c - 1)c + n^2 + 1$.

iv) $\dim(\alpha^{-1}(0)) = (n^2 - c - 1)c + n^2 + c$.

Note that for fixed $A \neq 0$, we have $\alpha^{-1}(A) \cong \operatorname{Gr}(M_n(\mathbf{C})/\mathbf{C}A, n^2 - c - 1) \times Z(A)$, of dimension $(n^2 - c - 1)c + z_A$. Moreover, $\alpha^{-1}(0) \cong \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times M_n(\mathbf{C})$, of dimension $(n^2 - c)c + n^2 = (n^2 - c - 1)c + n^2 + c$. If $n < z < n^2$, then $0 \not\in \Omega_z$, so it follows that $\dim \alpha^{-1}(\Omega_z) = \dim \Omega_z + (n^2 - c - 1)c + z$, and ii) follows by Claim 2. Points iii) and iv) follow similarly, as does the dimension calculation of i).

So it remains to prove that $\alpha^{-1}(\Omega_n)$ is irreducible. For this, since $\Omega_n$ is irreducible and $\alpha^{-1}(\Omega_n) \to \Omega_n$ has irreducible fibers, it suffices to show that this map admits sections locally on $\Omega_n$. So fix $A \in \Omega_n$ and let $V$ be a codimension $c + 1$ subspace of $M_n(\mathbf{C})$ not containing $A$. Let $W$ be a neighborhood of $A$ in $\Omega_n$ such that $A' \not\in V$ for all $A' \in W$, and note that there is a section $\sigma\colon W \to \alpha^{-1}(W)$ defined by $\sigma(A') = (V + \mathbf{C}A', A', 0)$.

Claim 4. If $X$ is an irreducible component of $\mathcal{D}$, then either $X$ is the closure of $\alpha^{-1}(\Omega_n)$ or $X = \alpha^{-1}(0)$ and $c = n$.

We have $X = \bigcup_{z = n}^{n^2} (X \cap \alpha^{-1}(\Omega_z))$, so there is a unique $z$ such that $X \cap \alpha^{-1}(\Omega_z)$ is open in $X$. In particular, we have $\dim X \leq \dim \alpha^{-1}(\Omega_z)$. If $n < z < n^2$, then Claim 3 ii) shows that $\dim X < (n^2 - c - 1)c + n^2 + n$, contradicting Claim 1. If $z = n^2$, then one of $X \cap \alpha^{-1}(\Omega_{n^2} - \{0\})$ and $X \cap \alpha^{-1}(0)$ is open in $X$. The former case cannot occur for the same reason as above (using Claim 3 iii)), while by Claim 1 the latter case can only occur if $\dim \alpha^{-1}(0) \geq (n^2 - c - 1)c + n^2 + n$. Since $c \leq n$, Claim 3 iv) shows that in this case $c = n$, and we see $X = \alpha^{-1}(0)$. Finally, if $z = n$ then $X$ is contained in the closure of $\alpha^{-1}(\Omega_n)$, and by Claim 3 i) this containment is an equality.

Proof of Proposition. If $c < n$, then Claim 3 i) and Claim 4 show that $\mathcal{D}$ is irreducible of the claimed dimension. If $c = n$, then Claim 3 i), iv) and Claim 4 show that $\mathcal{D} = \alpha^{-1}(0) \cup \overline{\alpha^{-1}(\Omega_n)}$, with both irreducible components of the claimed dimension. QED