Here is what I know about this (and mostly learned from my student Thilo Weinert):
It is known that $\mathbb Q\not\rightarrow(\omega+1,4)^3$.
I.e., there is a coloring of the unordered triples of rationals with two colors $0$ and $1$
such that no set of rationals of ordertype $\omega+1$ is homogeneous of color $0$ and
no set of size 4 is homogeneous of color 1.
On the other hand, Milner and Prikry showed $\omega_1\rightarrow(\omega\cdot 2,4)^3$
and Jones proved $\omega_1\to (\omega+m,n)^3$ for all $n,m\in\omega$.
Milner and Prikry conjectured that $\omega_1\rightarrow(\alpha,n)^3$ holds for all
$n\in\omega$ and all countable $\alpha$.
Note that all the positive results are unsymmetric, i.e., they only give finite homogeneous sets in one of the colors.
Also, there is a definable (continuous, actually) coloring of the unordered triples of $2^\omega$ (ordered lexicographically) with two colors such that
every infinite homogeneous set has ordertype either $\omega$ or $\omega^*$.
Finally, let us consider this coloring: for each $\alpha\lneq\omega_1$ fix a wellordering $\leq_\alpha$ of $\alpha$ of ordertype $\leq\omega$.
Given three ordinals $\alpha\lneq\beta\lneq\gamma\lneq\omega_1$, let $c(\alpha,\beta,\gamma)=0$ if $\leq_\alpha$ agrees with the usual ordering of ordinals
on $\{\alpha,\beta\}$ and otherwise let $c(\alpha,\beta,\gamma)=1$.
I think there is no homogeneous set of color $0$ of ordertype $\omega+2$ and no
homogeneous set of color $1$ of ordertype $\omega+1$. This would show $\omega_1\not\rightarrow(\omega+2)^3_2$.
