This depends on how strictly one interprets the word 'analogues'.

For example, there cannot be any inequality on the Ricci tensor alone that is equivalent to positive sectional curvature in dimensions higher than $3$. The reason is that such a condition would then imply that every Einstein metric of positive scalar curvature in dimension greater than $3$ would have positive sectional curvature (since the criterion would have to hold for the round spheres), and this is known to be false. In fact, in higher dimensions, the irreducible symmetric spaces of compact type are Einstein with positive scalar curvature, but most of them have some sectional curvatures vanishing. Moreover, it would not be hard to construct Einstein metrics in dimension $4$ with positive scalar curvature that have some negative sectional curvatures. For example, many Kähler-Einstein metrics in dimension $4$ with positive scalar curvature do have some negative sectional curvatures.

On the other hand, in dimension $n$, the space of Riemann curvature tensors with positive sectional curvature is an open cone on the space of Riemann curvature tensors and hence it can be described by some set of (strict) inequalities that are invariant under the natural action of $\mathrm{O}(n)$. When $n>3$, these inequalities will have to involve not only the Ricci curvature but the Weyl curvature as well. If you are willing to consider these as analogues of the inequality that works in dimension $3$ (where the Weyl curvature vanishes identically), then they answer would be 'yes, there is an analogue in each dimension'.

This depends on how strictly one interprets the word 'analogues'.

For example, there cannot be any inequality on the Ricci tensor alone that is equivalent to positive sectional curvature in dimensions higher than $3$. The reason is that such a condition would then imply that every Einstein metric of positive scalar curvature in dimension greater than $3$ would have positive sectional curvature (since the criterion would have to hold for the round spheres), and this is known to be false. In fact, in higher dimensions, the irreducible symmetric spaces of compact type are Einstein with positive scalar curvature, but most of them have some sectional curvatures vanishing. Moreover, it would not be hard to construct Einstein metrics in dimension $4$ with positive scalar curvature that have some negative sectional curvatures. For example, many Kähler-Einstein metrics in dimension $4$ with positive scalar curvature do have some negative sectional curvatures.

On the other hand, in dimension $n$, the space of Riemann curvature tensors with positive sectional curvature is an open cone on the space of Riemann curvature tensors and hence it can be described by some set of (strict) inequalities that are invariant under the natural action of $\mathrm{O}(n)$. When $n>3$, these inequalities will have to involve not only the Ricci curvature but the Weyl curvature as well. If you are willing to consider these as analogues of the inequality that works in dimension $3$ (where the Weyl curvature vanishes identically), then the answer would be 'yes, there is an analogue in each dimension'.