algebraic cycles - Geometry of the multilagrangian Grassmannian

This is a partial answer to your question: I believe there is exactly one class not coming from $G(3,6)$, and that this class will be represented by an algebraic cycle. However, I don't have a description of it.

My reasoning: Let's count points on your variety over a field with $q$ elements. I get
$$1+q+2q^2+3 q^3+ 4 q^4 + 3q^5 + 2 q^6 + q^7 + q^8.$$
(More details below.)
I have checked that your variety is smooth, so the Weil conjectures tell us that $H^4$ is four dimensional. Three of those dimensions come from $H^4(G(3,6))$, leaving one unexplained. Moreover, all the classes from $G(3,6)$ lie in $H^{2,2}$, so $H^{2,2}$ of your variety has dimension at least $3$. But then, by the symmetry of the Hodge diamond, the missing class is also in $H^{2,2}$. Assuming the Hodge conjecture, some multiple of it must be an algebraic class.

The point count: Write a point in your variety as the row span of a $3 times 6$ matrix $left( A B right)$. Your multi-Lagrangian condition is that $det A=det B$. This means either $A$ and $B$ both have rank $3$, or neither does. But your matrix is required to have full rank, so the ranks of $A$ and $B$ can be either $(3,3)$, $(2,2)$, $(2,1)$ or $(1,2)$. Remember also that we are counting not just $3 times 6$ matrices obeying these conditions, but orbits of such matrices under the left action of $GL_3$. I get

Rank $(3,3)$: $(q^3-1)(q^3-q)q^2$ points.
Rank $(2,2)$: $(q^2+q+1)(q^2+q+1)(q^3+q^2-q-1)$ points.
Rank $(2,1)$: $(q^2+q+1)(q^2+q+1)$ points.
Rank $(1,2)$: $(q^2+q+1)(q^2+q+1)$ points.

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The following is the record of an incorrect solution, left as a warning to others. Let $A$ be the subvariety of all three planes which are of the form $u wedge v wedge w$ where $u$ and $v$ are in $mathrm{Span}(e_1, e_2, e_3)$ and $w$ is in $mathrm{Span}(e_4, e_5, e_6)$. Let $B$ be the similar subvariety where I switch the roles of $123$ and $456$.

It is easy to show that $[A]-[B]$, in $H^4(MG(3,6))$, is orthogonal to the classes obtained by restriction from $G(3,6)$. This would suggest that $[A]-[B]$ is our missing class. Unfortunately, it turns out that $[A]-[B]=0$. Proof: Let $S$, in $G(3,6)$, be the locus of those $3$-planes which meet $mathrm{Span}(e_1,e_2,e_3)$ in dimension $2$. Then $S cap MG(3,6)=A$, and the intersection is transverse. Let $T$ be those $3$-planes which similarly meet $mathrm{Span}(e_4,e_5,e_6)$; so $B=T cap MG(3,6)$. But $S$ and $T$ are homologous in $G(3,6)$, so $A$ and $B$ are homologous in $MG(3,6)$. Grrrr...