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Let's look at the statement of Rice's theorem

Suppose we have a set of languages $\mathcal{S}$. Then the problem of deciding whether the language of a given Turing machine is in $\mathcal{S}$ is undecidable, provided that there exists a Turing machine that recognizes a language in $\mathcal{S}$ and a Turing machine that recognizes a language not in $\mathcal{S}$.

Thus the theorem is not concerned with how "easy" the property of languages may be, but how hard it is to recognize even easy properties of a language $L$, given only a Turing machine that generates $L$. It says that the only recognizable sets of languages are the empty set (recognized by saying NO for every Turing machine) and the set of all languages (recognized by saying YES for every Turing machine).

This is not so surprising: since you can't in general tell whether a Turing machine halts, you really don't know anything about what it ultimately does.

Addendum. Your intuition that Rice's theorem shows the halting problem to be "easiest" is correct if you amend your statement as follows: the halting problem is the easiest in the class of problems $P_{\mathcal S}$, where $P_{\mathcal S}$ is the problem of deciding, for an arbitrary Turing machine $M$, and nontrivial language set $\mathcal{S}$, whether the language generated by $M$ is in set $\mathcal{S}$.

The reduction in the proof of Rice's theorem shows that every such problem $P_{\mathcal S}$ is at least as hard as the halting problem, and one can point to sets $\mathcal{S}$ for which the problem $P_{\mathcal S}$ is actually harder, for example, the set of infinite languages.

Further exploration of the problems harder than the halting problem leads you the study of degrees of unsolvability.

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Let's look at the statement of Rice's theorem

Suppose we have a set of languages $\mathcal{S}$. Then the problem of deciding whether the language of a given Turing machine is in $\mathcal{S}$ is undecidable, provided that there exists a Turing machine that recognizes a language in $\mathcal{S}$ and a Turing machine that recognizes a language not in $\mathcal{S}$.

Thus the theorem is not concerned with how "easy" the property of languages may be, but how hard it is to recognize even easy properties of a language $L$, given only a Turing machine that generates $L$. It says that the only recognizable sets of languages are the empty set (recognized by saying NO for every Turing machine) and the set of all languages (recognized by saying YES for every Turing machine).

This is not so surprising: since you can't in general tell whether a Turing machine halts, you really don't know anything about what it ultimately does.