Let me start with a simple observation. Suppose $f$ is a weight two newform of level $p^3$. Write $d$ for the size of the Galois orbit $f^\sigma, \sigma \in \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$. Then $d \geq (p-1)/2$. The proof is quite simple: associated to $f$ is an abelian variety $A\_f$ of dimension $d$ and conductor $p^{3d}$. If a prime $p$ divides the conductor of an abelian variety of dimension $d$, and $p>2d+1$, then the maximal power of $p$ dividing the conductor is $p^{2d}$ (I believe this is a theorem of Serre-Tate). Hence if we had $d < (p-1)/2$, this would immediately imply the absurd inequality $3d<2d$, so contradiction. Of course this also works for newforms of weight two and higher prime power level.


My questions:


1. Is this written down in the literature somewhere? It seems like such a simple observation that I want to guess that it is, but I cannot find a reference.
2. Is the same result true for higher weight newforms?



Galois orbits of newforms with prime power level
See Section 4 of my paper [The Hecke algebra T\_k has large index](http://www.math.northwestern.edu/~emerton/pdffiles/index.pdf), joint with Frank Calegari, where this sort of argument is applied. Lemma 4.1 is a variant of the statement you prove; see also the last sentence. 


In any case, the argument does generalize to higher weight newforms; it is just a statement
about $p$-torsion elements in $GL\_n(\mathbb Z\_p)$ (there are none when $p > n + 1$). This
in turn implies that a Galois rep. into $GL\_n(\mathbb Z\_p)$ must be tamely ramified at $p$
if $p > n + 1$, and hence has conductor at $p$ bounded by $p^n$ (because the exponent of the conductor is then just the codimension of the fixed subspace under tame inertia). Note also
(see our paper) that you can replace $\mathbb Z\_p$ by any unramified extension, and thus
that the argument controls not only the degree of the coefficient field of a newform, but also the ramification of $p$ in the coefficient field (the larger the degree of the coeff.
field, the more ramified $p$ has to be). 


