For a quantum mechanical description of a system (like a small molecule) we can write:
$$\langle\psi|\hat {H}|\psi\rangle = \overline E$$
Question:
Is that energy the same as zero Kelvin energy obtained by statistical mechanics (using $E_n$ energies and partition functions)?
The average $\langle E\rangle$ you have on the right is not to be understood in the sense of statistical mechanics (and thus as a temperature).
The wavefunction $\psi$ can be a linear combination of states $\psi_n$ of definite energy of your molecule. The $\psi_n$’s are solutions of the Schrodinger equation. Thus, \begin{align} \psi(x,t)&=\sum_i a_i \psi_i(x,t)\, ,\qquad \sum_i \vert a_i\vert^2=1\, , \tag{1} \\ \hat H\psi_i(x,t)&=E_i\psi_i(x,t)\, . \end{align} With this wavefunction the average energy $$ \langle E\rangle = \sum_i \vert a_i\vert^2 E_i \tag{2} $$ is a weighted average of the possible energies of your molecule, with the modulus square $\vert a_i\vert^2$ of the complex amplitude $a_i$ functioning as statistical weight. $\psi(x,t)$ given in (1) is an example of a pure state.
For instance, if you have a hydrogen wavefunction of the form $$ \psi(r)=a_1\psi_{100}(r)+a_2\psi_{200}(r) $$ then the average energy of this system is $$ \vert a_1\vert^2\times (-13.6) + \vert a_2\vert^2 \times (-13.6/4). \tag{3} $$ Measuring the energy, you will sometimes get the value $-13.6$ (this outcome will occur $\vert a_1\vert^2$ of the time) and sometimes get the value $-13.6/4$ (this outcome will occur $\vert a_2\vert^2$ of the time). The average energy is exactly given by (3).
In quantum statistical mechanics one introduces the concept of a mixted state. For mixed states one cannot define a wavefunction for the system as in (1). The system is described using a matrix that represents a statistical mixture of wavefunction that is not of the form given in (1); the average energy for this statistical mixture is also not of the form (2) since the latter comes from (1), which does not exist for mixted states.
Statistical mechanics does not really apply to the kind of system your equation refers to: your equation is good for a pure quantum state $\psi$. In statistical mechanical terms, the system's microstate is exactly and fully specified by $\psi$, and you can't really therefore talk about a system's temperature when it is in a pure state.
Recall that pure states can be non ground states, or in a superposition containing non ground states. So there is, in general, no relationship definable with a "$0{\rm K}$ energy".
A classical mixture of pure quantum states can describe a statistical mechanical system, can therefore be assigned a temperature and is described by a density matrix $\rho$. In such a case, your equation becomes replaced by:
$$\langle E\rangle = \mathrm{tr}(\rho\,\hat{H})$$
Perhaps a kind of converse to your question makes more sense and yields a more concrete answer: as we cool a statistical mechanical system down, and if the ground state is nondegenerate, the system becomes "forced" into a state where all particles / ensemble members are in the ground state. If, further, the ground state is nondegenerate (i.e. there is only one ground quantum energy eigenstate), then the $0{\rm K}$ state is a multiparticle pure quantum state and this state is the ground energy eigenstate. User Joshphysics analyzes this behavior in more detail in his answer here.
