If $K$ is an algebraically closed field whose characteristic doesn't divide $n$, then the Galois theory of this extension is not complicated. Indeed $K(E)/[n]^* K(E)$ is then abelian with Galois group canonically isomorphic to $E[n]$. A point $P \in E[n]$ acts by translation on $K(E)$, namely $\sigma_P(f) = t_P^* f$ where $t_P : E \to E$ is the translation map, and the $\sigma_P$ are exactly the elements of the Galois group. A function $f \in K(E)$ will generate the extension if and only if its translates by $E[n]$ are pairwise distinct. It is not hard to show that, taking for example a Weierstrass model for $E$, the function $f=x$ works.

In the case $K$ is not algebraically closed one has to be careful that the extension isn't necessarily Galois anymore (it is if and only if the $n$-torsion is rational over $K$). However, regarding the question about generators, the above criterion generalizes, namely $f \in K(E)$ generates the extension if and only if its translates (which lie in $K_n(E)$, where $K_n$ is the field obtained from $K$ by adjoining the coordinates of the $n$-torsion points) are pairwise distinct. So, for example, $f=x$ is still a generator of the extension.

**EDIT.** In view of the OP's comments, here is an alternate way of constructing generators of the extension $K(E)/[n]^* K(E)$. These generators will have minimal polynomials of the form $X^n-[n]^* g$, for reasonable functions $g \in K(E)$.

Assume $E[n] \subset E(K)$, so that $K(E)/[n]^* K(E)$ is Galois and $\operatorname{Gal}(K(E)/[n]^* K(E)) \cong E[n]$, as explained above. By Kummer theory, this extension is generated by two $n$-th roots of suitable elements of $[n]^* K(E)$. We can find these elements as follows. Let $f \in K(E)$ such that $f^n = [n]^* g$ for some $g \in K(E)$. Taking the divisors, we see that $\operatorname{div}(f)$ is invariant by translation by $E[n]$, so it has the form

\begin{equation*}
\operatorname{div}(f) = \sum_i \sum_{R \in E[n]} \lambda_i [P_i+R]
\end{equation*}
with $\lambda_i \in \mathbf{Z}$ and $P_i \in E$. Conversely, such a divisor is principal if and only if $\sum \lambda_i = 0$ and $n^2 \sum \lambda_i P_i = 0$ (because the sum of all $n$-torsion points is zero). The divisor of $g$ is then given by

\begin{equation*}
\operatorname{div}(g) = n \sum_i \lambda_i [n P_i]
\end{equation*}
Note that if $n \sum \lambda_i P_i = 0$ then $f$ is already in $[n]^* K(E)$. So, letting $Q_i=nP_i$, the conditions on the divisor of $g$ are $\sum \lambda_i = 0$ and $\sum \lambda_i Q_i \in E[n] \backslash \{0\}$.

For example, for each $Q \in E[n]$, one can take $g_Q$ such that $\operatorname{div}(g_Q)=n[Q]-n[0]$. Identifying the base field of the extension with $K(E)$ and denoting the extension by $L/K(E)$, we thus get $L=K(E)(g_{Q_1}^{1/n},g_{Q_2}^{1/n})$ if and only if $(Q_1,Q_2)$ is a basis of $E[n]$. Note : the Galois action on these functions $g_{Q}^{1/n}$ is related to the Weil pairing (see for example Silverman, *The arithmetic of elliptic curves*). More generally one can define $g_D$ for any $D \in I$, where $I$ is the augmentation ideal of $\mathbf{Z}[E[n]]$. Then $L=K(E)(g_{D_1}^{1/n},g_{D_2}^{1/n})$ if and only if the classes of $D_1$ and $D_2$ generate $I/I^2 \cong E[n]$. Finally, there is the further flexibility in that one can choose functions $g$ whose divisors are not necessarily supported on $E[n]$.

division polynomials, see en.wikipedia.org/wiki/Division_polynomials. Moreover, the roots of the $(2n + 1)$th division polynomial $\psi_{2n + 1}$ are exactly the $x$ coordinates of the points of $E[2n+1] \setminus O$. – Francesco Polizzi Nov 10 '11 at 16:30