What is known about the plethysm $\text{Sym}^d(\bigwedge^3 \mathbb{C}^6)$

No, it is not multiplicity-free. Already for $d=6$, this representation contains the Schur functor $S^{4,4,4,2,2,2}$ twice. This can be easily checked in Magma (even the online calculator) issuing the commands

Q := Rationals();

s := SFASchur(Q);

s.[6]~s.[1,1,1];

Ignoring the Schur functors that vanish on $\mathbb{C}^6$, we obtain

s.[4,3,3,3,3,2] + 2*s.[4,4,4,2,2,2] + s.[5,4,3,3,2,1] + s.[5,4,4,2,2,1] + s.[5,5,4,2,1,1] + s.[5,5,5,1,1,1] +s.[6,3,3,3,3] + s.[6,4,4,2,2] + s.[6,5,5,1,1] +s.[6,6,6]

A trivial remark: for $d=4k$ the module $[2k,2k,2k,2k,2k,2k]$ should appear. This is because of the degree four relative invariant listed in Proposition 7 page 81 of the classic and very long paper by Sato and Kimura on Prehomogeneous vector spaces.

The quartic invariant is described explicitly in Ring of invariants of $\operatorname{SL}_6$ acting on $\Lambda^3 \mathbb C^6$

Robert Bryant mentions in that question that the quartic invariant generates the invariant ring. This should imply that the only modules $[m,m,m,m,m,m]$ you will see for general $d$ are the ones I just described, with multiplicity one.

$$\bigoplus_{k \ge 0} t^k Sym^{k} V = \frac{1}{(1−tV)(1−t^2 \mathbf{g})(1−t^3 V)(1−t^4)(1−t^4 V_2)},$$ where $V = \wedge^3 \mathbb{C}^6 = [0,0,1,0,0], V_2 = [0,1,0,1,0]$, and $\mathbf{g} = [1,0,0,0,1].$

See section 6 of "Series of Lie Groups" by Landsberg and Manivel.