How do the Dim($K_f / \mathbf{Q}$) vary for all f in a given $S_k(\Gamma(N))$?

Question

For the theory of classical modular forms, the space of new forms $S_k^{new}(\Gamma(N))$ has a basis of Hecke eigenforms $\{ f_i = \sum a_n q^n : a_1=1, a_n \in \bar{\mathbb{Q}}\}$

Given $k$ and $N$ (I'll take answers restricted to $\Gamma_0(N)$ or $\Gamma_1(N)$)

1) can we say anything about $\{dim_{\mathbf{Q}}[K_{f_i} : \mathbf{Q}] : f_i \}$, where $K_{f_i} = \mathbf{Q}[ a_n : n \geq 1 ]$? (Are they all equal? different? Are the fields themselves equal?)

2) Can we know an explicit $L = \cup K_{f_i}$ apriori? (My guess is that it's related to the defining equation of the projective curve $X(\Gamma)$).

We know that, for each $i$, the dimension of $K_{f_i}$ is finite, but what can we say about all of the $f_i$ collectively?

Simplified Question: I know that, via the modularity of rational elliptic curves, If $E/\mathbf{Q}$ is an elliptic curve of conductor N, then it's associated modular form $f \in S_2(\Gamma_0(N))$ has $dim_{\mathbf{Q}}[K_{f}:\mathbf{Q}] = 1$. Is it true that all of the $f_i$, new eigenforms, in $S_2(\Gamma_0(N))$ have $K_f = \mathbf{Q}$?

Answer 1

The answer to your last question is no. In general, if $K_f/\mathbb Q$ has degree $d$, then there is an associated factor of $J_0(N)$ defined over $\mathbb Q$ of dimension $d$. This is all in Shimura's Arithmetic Theory of Automorphic Forms, I think.

For your question (2), the curves $X_0(N)$ and $X_1(N)$ are defined over $\mathbb Q$, so their fields of definition have nothing to do with the field that you're calling $L$.

Finally, it is known that the number of isogeny classes of elliptic curves of conductor $N$ is less than $O(N^{1/2+\epsilon})$ (and results of Pierce, Helfgott, and Venkatesh reduce the exponent at bit). Since each such isogeny class appears as an elliptic factor of $J_0(N)$ (thanks to Wiles et.al.), and each elliptic factor gives a cusp form with coefficients in $\mathbb Q$, it follows that there are at most $O(N^{1/2+\epsilon})$ normalized cusp forms with coefficients in $\mathbb Q$. But the total number of normalized cusp forms with $\overline{\mathbb Q}$ coefficients is the genus of $X_0(N)$, which is "roughly" $N/12$. Conclusion: Most normalized cusp forms for $\Gamma_0(N)$ do not have coefficients in $\mathbb Q$.