It's not always easy to find in the various references, but there is a fairly explicit criterion to check this on a Riemannian $4$-manifold that does not require that one actually compute the isotropic curvature in every isotropic $2$-plane. For example, while it is not obvious from the definition, the isotropic curvature only depends on the scalar curvature and the Weyl curvature; the traceless Ricci curvature does not enter into the final formula.
For simplicity, assume that $M^4$ is oriented (though the final answer will not depend on the orientation). Then the $2$-form bundle $\Lambda^2(TM)$ splits into two $3$-plane bundles $\Lambda^2_\pm(TM)$ of self-dual ($+$) and anti-self-dual ($-$) bivectors. Then the Riemann curvature $R:\Lambda^2\to\Lambda^2$ has the Singer-Thorpe decomposition $$ R = \left(\frac{S}{12}\right)\,\mathrm{id} + W_+ + W_- + Q + Q^* $$ where $S$ is the scalar curvature, the symmetric maps $W_\pm:\Lambda^2_\pm\to\Lambda^2_\pm$ are traceless, and $Q:\Lambda^2_+\to\Lambda^2_-$. (In this formulation $W_\pm$ are the self-dual and anti-self-dual components of the Weyl curvature while $Q$ is equivalent to the traceless Ricci curvature.)
In these terms, letting $\lambda(W_\pm)\ge 0$ denote the (pointwise) maximum eigenvalue of $W_\pm$ as a function on $M$ (nonnegative because the trace of $W_\pm$ vanishes), it turns out that the metric has nonnegative isotropic curvature if and only if $$ S\ge 6\lambda(W_\pm) $$ Note that, because this condition is unchanged by reversing the orientation of $M$, it is actually well-defined even for unoriented Riemannian $4$-manifolds.
Positive isotropic curvature is equivalent to $S>6\lambda(W_\pm)$.