The $\alpha$=0 point on the polar would be a poor metric - there is nothing special about it, and your aircraft will not fly at this polar point when you strive for longest endurance. But your common sense is correct in predicting that this will be a slow airplane.
The precise answer normally depends on your means of propulsion. Since this is for gliders, the answer will be quite simple.
Your definition of efficiency is minimum energy loss. Energy means potential energy in this case $E_{pot} = m\cdot g\cdot h$, and the loss of potential $\frac{dh}{dt}$ energy over time is expressed as the sink speed $v_s$. So we need to find the polar point at which the glider will have the lowest possible sink speed.
Let's start with the parabolic drag equation which splits the total drag coefficient $c_D$ in one component which is constant over the lift coefficient range and one which changes with the square of the lift coefficient.
$$c_D = c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$
The symbols are:
$\kern{5mm} m\:\:\:\:$ aircraft mass
$\kern{5mm} g\:\:\:\:\;$ gravitational acceleration
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} v_z\:\:\;$ sink speed
$\kern{5mm} c_{D0} \:$ zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$ lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$ 3.14159$\dots$
$\kern{5mm} AR \:\:$ aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$ the wing's Oswald factor
$\kern{5mm} S \:\:\:\:\:$ wing area
Next, we need an expression for the sink speed. This should be easy: It is the forward speed times the tangent of the glide path angle $\gamma$. If you allow to approximate the tangent function for small angles by the radian of the angle, you can write:
$$v_z = v\cdot tan\gamma ≈ v\cdot\gamma = v\cdot\frac{c_D}{c_L}$$
Now insert the drag coefficient
$$v_z = v\cdot\left(\frac{c_{D0}}{c_L} + \frac{c_L}{\pi\cdot AR\cdot\epsilon}\right)$$
and make the speed dependency of the lift coefficient obvious:
$$v_z = \frac{c_{D0}\cdot\frac{\rho\cdot v^3}{2}\cdot S}{m\cdot g} + \frac{m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\frac{\rho\cdot v}{2}\cdot S} = v^3\cdot\frac{c_{D0}\cdot\rho\cdot S}{2\cdot m\cdot g} + \frac{1}{v}\cdot\frac{2\cdot m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\rho\cdot S}$$
Now we are ready to differentiate with respect to $v$ and find the condition when the derivation will become zero.
$$\frac{dv_z}{dv} = 3\cdot v^2\cdot\frac{c_{D0}\cdot\rho\cdot S}{2\cdot m\cdot g} - \frac{1}{2\cdot v^2}\cdot\frac{2\cdot m\cdot g}{\pi\cdot AR\cdot\epsilon\cdot\rho\cdot S}$$
Now re-insert the lift coefficient, which simplifies the equation mightily:
$$\frac{dv_z}{dv} = 3\cdot\frac{c_{D0}}{c_L} - \frac{c_L}{\pi\cdot AR\cdot\epsilon}\,\overset{!}{=}\, 0$$
$$\Rightarrow 3\cdot c_{D0} = \frac{c^2_L}{\pi\cdot AR\cdot\epsilon}$$
So your energy loss will be minimized when the induced drag is three times bigger than the zero-lift drag. The polar point is then:
$$c_L = \sqrt{3\cdot c_{D0}\cdot \pi\cdot AR\cdot\epsilon}\;\;\text{and}\;\;c_D = 4\cdot c_{D0}$$
With wings of a high aspect ratio this gives a rather high lift coefficient, so you might need to pick a high-lift airfoil in order to actually trim this polar point. The figure of merit when selecting the airfoil should be the ratio $\frac{c^{³⁄₂}_L}{c_D}$; this should reach its maximum at the lift coefficient of minimum energy loss. Enter the lift and drag coefficients into a spreadsheet and create a new column for $\frac{c^{³⁄₂}_L}{c_D}$ (or $\frac{c^3_L}{c_D^2}$; doesn't matter). Select an airfoil that maximises this value at the calculated optimum lift coefficient.
