The answer is that the relation $C(\phi)$ defined as "every Turing ideal satisfies $\phi$" is $\Pi^1_1$ complete.
To obtain a lower bound, it's enough to show that a $\Pi^1_1$ sentence of second-order arithmetic is true if and only if it is true in every Turing ideal. Half of this equivalence follows from the fact that the standard model is a Turing ideal, so anything false is false in that ideal. Conversely, any Turing ideal in which a $\Pi^1_1$ statement is false yields a counterexample to the statement which is still a counterexample in the standard model. The key point is that any arithmetical property of a real is absolute to all Turing ideals, including the standard model. Thus the satisfaction relation for $\Pi^1_1$ formulas agrees with the $C$ relation, and so the latter is $1$-reducible to the former.
To get the matching upper bound we have to show that $C(\phi)$ can be defined by a $\Pi^1_1$ formula. To do this, we need a little lemma.
Lemma. If a sentence of second-order arithmetic is false in any Turing ideal then it is false in a countable Turing ideal.
To prove this, suppose $\phi$ is false in an ideal $I$. View $I$ as an $\omega$-model in the language of second-order arithmetic. By the downward Lowenheim-Skolem theorem, there is a countable elementary substructure $S$ of $I$. Then $S$ will be an $\omega$-model because it is a substructure of an $\omega$-model, and $S$ will be a Turing ideal, because this property of an $\omega$-model is definable by a formula of second-order arithmetic that was true in $I$. And $\phi$ will be false in $S$, again because it was false in $I$.
Thus, in light of the lemma, a sentence of second-order arithmetic is true in every Turing ideal if and only if it is true in every countable Turing ideal.
Now consider the formula $C'(\phi)$ which says "every countable Turing ideal satisfies $\phi$". The relation $C'$ can be expressed in second-order arithmetic as "For every $X$, for every $Y$, if $X$ codes a Turing ideal and $Y$ is the satisfaction predicate for $X$ then $Y(\phi) = 1$". Here "$X$ codes a Turing ideal" means that the class $\{ \{ m: 2^n3^m \in X\} : n \in \omega\}$ is a Turing ideal. . The two hypotheses of the if statement in the definition are both expressible by arithmetical formulas with parameters for $X$ and $Y$, so the entire formula is $\Pi^1_1$.
Hence the $C$ relation is $1$-reducible to the satisfaction relation for $\Pi^1_1$ sentences. Thus, by Myhill's theorem, $C$ is $1$-equivalent to that satisfaction relation.