It's a bit unclear to me what conventions you're using to identify $\mathfrak{u}(n)^*$ with $Herm(n)$ - it seems like you're taking $\langle X, Y\rangle = i\operatorname{Tr}(XY)$ or something. But I'm going to use $\langle X,Y\rangle = \operatorname{Tr}(XY)$ to identify $\mathfrak{u}(n)^*$ with $\mathfrak{u}(n)$, the latter being the anti-Hermitian $n\times n$ matrices (so we're working on the adjoint orbit in $\mathfrak{u}(n)$).
First, to correct one thing from your question: the symplectic form is
$$
\omega_X(\dot{X}_1,\dot{X}_2) = \langle X, [\Omega_1,\Omega_2]\rangle.
$$
The expression you give holds when $X=D$. I'm now taking $D = \operatorname{diag}(i\lambda_1, \ldots, i\lambda_n)$ where the $\lambda_i$ are all real.
In general, if $\mathfrak{g}$ is a simple, real and compact Lie algebra, then $\operatorname{ad}_X$ for $X\in\mathfrak{g}$ will have pure imaginary eigenvalues, with corresponding eigenvectors in the complexification $\mathfrak{g}^\mathbb{C}$ of $\mathfrak{g}$. Let $\mathfrak{n}^0_X, \mathfrak{n}^+_X, \mathfrak{n}^-_X$ denote the span of the eigenvectors corresponding to zero, positive, and negative imaginary axis eigenvalues respectively. Note that $\mathfrak{n}_X^0 = \mathfrak{g}_X^\mathbb{C}$, the complexification of the stabiliser algebra of $\mathfrak{g}$ at $X$. Define
$$
P_X:= \lbrace \operatorname{ad}_XY \mid Y\in \mathfrak{n}_X^-\rbrace.
$$
Then it's not too difficult to see that $P$ is an $\operatorname{Ad}_G$-invariant totally complex polarization for the adjoint orbit $\mathcal{O}\subset \mathfrak{g}$ through $X$, meaning that $P$ is integrable, Lagrangian wrt $\omega$, and $P_X \oplus \overline{P_X} = (T_X\mathcal{O})^\mathbb{C}$. The corresponding complex structure $J$ on $\mathcal{O}$ multiplies $P$ by $+i$, and $\overline{P}$ by $-i$.
This construction is equivalent to the one described in the linked answer in your question: there you take a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}^\mathbb{C}$ containing $X$, and take the root-space decomposition
$$
\mathfrak{g}^\mathbb{C} = \mathfrak{h} \oplus \bigoplus_{\alpha\in \Delta} R_\alpha.
$$
Then in my notation above
$$
\mathfrak{n}_X^0 = \mathfrak{h}\oplus\bigoplus_{\alpha(X)=0}R_\alpha\quad\text{and}\quad
\mathfrak{n}_X^\pm = \bigoplus_{\alpha(X)\in i\mathbb{R}_{\pm}} R_\alpha.
$$
Applying the above construction to the case $\mathfrak{g} = \mathfrak{u}(n)$, and taking $D = \operatorname{diag}(i\lambda_1, \ldots, i\lambda_n)$, the eigenvectors of $\operatorname{ad}_D$ are the matrices $E_{pq}$ with a 1 in the $pq$ entry, and $0$ elsewhere
$$
[D,E_{pq}] = i(\lambda_p-\lambda_q)E_{pq}.
$$
If we further assume that $\lambda_i>\lambda_j$ for $i>j$, then $\mathfrak{n}_D^0$ is just all the diagonal $n\times n$ complex matrices, and $\mathfrak{n}_D^\pm$ are the lower/upper triangular matrices. Then
$$
J_D(\operatorname{ad}_D(E_{pq})) = i\operatorname{sign}(q-p)\operatorname{ad}_D(E_{pq}).
$$
The Kaehler metric at $D$ (extended by linearity to the complexified tangent space) is given by
$$
g_D(\operatorname{ad}_D(E_{pq}),\operatorname{ad}_D(E_{rs})) = \omega_D(\operatorname{ad}_D(E_{pq}),J_D\operatorname{ad}_D(E_{rs})) = i\operatorname{sign}(s-r) \langle D, [E_{pq},E_{rs}]\rangle.
$$
This is non-zero only if $rs=qp$, in which case it becomes
$$
i\operatorname{sign}(p-q) \langle D, E_{pp}-E_{qq}\rangle = -\operatorname{sign}(p-q)(\lambda_p-\lambda_q) = -|\lambda_p-\lambda_q|.
$$
To find its value on real vectors, you need to take appropriate linear combinations of the $E_{pq}$,
$$
g_D(\operatorname{ad}_D(E_{pq}-E_{qp}),\operatorname{ad}_D(E_{pq}-E_{qp})) = g_D(\operatorname{ad}_D(iE_{pq}+iE_{qp}),\operatorname{ad}_D(iE_{pq}+iE_{qp})) = 2|\lambda_p-\lambda_q|\\
g_D(\operatorname{ad}_D(E_{pq}-E_{qp}),\operatorname{ad}_D(iE_{pq}+iE_{qp}))=0
$$
etc.
The expression for the metric is extended to arbitrary $X = \operatorname{Ad}_UD$ by $\operatorname{U}(n)$-invariance of all structures,
$$
g_X(\operatorname{ad}_X(Y_1), \operatorname{ad}_X(Y_2)) = g_D(\operatorname{ad}_D(\operatorname{Ad}_{U^{-1}}Y_1),\operatorname{ad}_D(\operatorname{Ad}_{U^{-1}}Y_2)).
$$
I've spent a good hour chasing signs in this answer, so you should check all my conventions carefully (they could be off in places).
