Write the left invariant Maurer-Cartan 1-form $\omega$ on $SO(n)$ as $$ \begin{pmatrix} \omega^i_j & \omega^i_J \\ \omega^I_j & \omega^I_J \end{pmatrix}. $$ The structure equations of Cartan are $d\omega=-\omega\wedge \omega$. I claim that the 1-form $\Omega = \sum_I \omega^I_1 \wedge \omega^I_2$ is a Kaehler form for the metric $g=\sum_{I,j} \omega^I_j \omega^I_j$. Clearly the linear algebra is ok: the 1-forms $\omega^I_j$ are semibasic for the map $SO(n) \to Gr(2,n)=SO(n)/SO(2)\times SO(n-2)$, and transform by orthogonal transformation under the action of the structure group, so $g$ is a metric on $Gr(2,n)$. The structure equations force $\Omega$ to be closed. The structure group action preserves the almost complex structure for which $\omega^I_1 + i\omega^I_2$ are complex linear, as (under right action $r_g$ for $g \in SO(2) \times SO(n-2)$, $r_g^* \omega = Ad(g)^{-1}\omega$, so if we write $$ g=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} $$ then $r_g^*\omega = b^{-1}ga$, so $$ r_g^*(\omega^I_1+i\omega^I_2)=\sum_J b^J_I(\omega^J_1+i\omega^J_2)(a_1 + i a_2). $$ This almost complex structure is complex, as $$ d(\omega^I_1+i\omega^I_2)=-(\omega^I_J-i\omega^1_2)\wedge(\omega^J_1+i\omega^J_2) $$ (by the structure equations) so there are no $(\omega^I_1+i\omega^I_2)\wedge(\omega^J_1-i\omega^J_2)$ terms, i.e. no torsion, i.e. the Nijenhuis tensor vanishes, so a complex structure.
As abx points out, if you take a 2-plane, say the span of an orthonormal basis $u,v \in \mathbb{R}^n$, then the vector $z=u+iv$ is null for the quadratic form $\sum z_i^2$, clearly. Conversely, every null vector arises this way uniquely up to a complex scalar. The 2-plane determines $z$ up to rescaling by a unit complex number, so $Gr(2,n)$ is a quadric hypersurface in $\mathbb{P}_{\mathbb{C}}^{n-1}$.