I would write it differently. First, consider the universal space $M$ consisting of tuples $(S,L_1,\dots,L_5)$, where $S$ is a cubic surface and $L_i$ are lines on $S$. This space lies inside the product $P^{19}\times Gr(2,4)^5$ and can be described as the zero locus of a canonical global section of the vector bundle
$$
E := O(1) \boxtimes (S^3U_1^* \oplus S^3U_2^* \oplus S^3U_3^* \oplus S^3U_4^* \oplus S^3U_5^*)
$$
on it, where $U_i$ is the tautological bundle on the $i$-th copy of the Grassmannian. Consequently, the structure sheaf of $M$ has the following Koszul resolution
$$
0 \to \Lambda^{20}E^* \to \dots \to \Lambda^2E^* \to E^* \to O \to O_M \to 0.
$$
We are interested in the image of $M$ in $Gr(2,4)^5$, so let us push forward the above resolution to $Gr(2,4)^5$. Note that $E^*$ restricts as a sum of $O(-1)$ to any fiber $P^{19}$ of the projection, hence the $p$-th term restricts as a sum of $O(-p)$. Thus restriction of almost all terms are acyclic, and they do not contribute to the pushforward. The only terms which do are the first and the last.
The first gives $O$, and the last gives
$$
\det(S^3U_1 \oplus \dots \oplus S^3U_5) = \otimes_{i=1}^5\det(S^3U_i) = O(-6,-6,-6,-6,-6).
$$
This means that the equation of the image of $M$ is a hypersurface of polydegree $(6,6,6,6,6)$ on the product of the Grassmannians. Of course its equation is the one given by David Speyer --- the determinant is a homogeneous polynomial of polydegree $(12,12,12,12,12)$ in coefficients of matrices, but if you write it in terms of the Plucker coordinates (which are quadratic in coefficients of matirces), it will have polydegree $(6,6,6,6,6)$.