So the graph has $\binom83 \binom52=560$ vertices and is regular of degree $4$? If so it might be clearer to say that the vertices are labelled by ordered pairs $({\{a,b,c\}},\{{d,e\}})$$(\{{a,b,c\}},\{{d,e\}})$. Then the vertices $({\{1,b,c\}},\{{d,8\}})$ give an independent set of size $60.$ Add the vertices $({\{b,c,8\}},\{{1,d\}})$ to get size $120$. I have no idea how good that is.
later I was going to comment that one could add the points $(\{{1,8,a\}},\{{b,c\}})$ and half the points $(\{{a,b,c\}},\{1,8\})$ to get $154.$ However Flo gets $180$ in a more elegant manner below. No independent set could have size $224$ as I'll show.
The graph is interesting. For convenience, color the edges so that a vertex $v=(\{{a,b,c\}},\{{d,e\}})$ is connected by a red edge to $w=(\{{f,g,h\}},\{{d,e\}})$ and by black edges to $(\{{a,b,c\}},\{{f,g\}}),(\{{a,b,c\}},\{{f,h\}})$ and $(\{{a,b,c\}},\{{g,h\}}).$ Ignoring the red edges divides the graph into $56$ connected graphs with $10$ vertices and regular of degree $3.$ These are quickly seen from the combinatorial description to be Peterson Graphs with indepence number $4.$ Hence the upper bound of $56 \cdot 4=224.$
Fixing the two element set gives $20$ points disjoint save for $10$ red edges. There are $11$ isomorphism classes of $10$ independent points (i.e. $10$ pairwise intersecting $3$-subsets of a $6$-set). The most symmetric has symmetry group the exceptional embedding of $A5$ in $S6.$ However an independent set of $224$ points would on average use $8$ out of each of those $10$ points so that is where I would look.