An object on an angular surface will fall off, when forces exerted along the slope (gravitational, tangent to the surface: gt) are greater then those pressing the object to the slope (gravitational, perpendicular to the slope, gp) or working in the opposite direction on the slope (friction, $fr$).
As an example, $10\ \mathrm{cm}$ of snow, on a 0.5-by-1-meter surface:
- $0.1 \times 0.5 \times 1 = 0.05\ \mathrm{m^3}$ of snow, weighing approximately $0.05 \ \mathrm{m^3}\times 300\ \mathrm{\dfrac{kg}{m^3}}$ (density of snow = $300\ \mathrm{\dfrac{kg}{m^3}}$) = $15\ \mathrm{kg}$.
- Gravitational force exerted = $9.81\ \mathrm{\dfrac{m}{s^2}} \times 15\ \mathrm{kg} = 147\ \mathrm{N}$.
- Frictional force $fr = 147\ \mathrm{N}\times 0.53$ (as a lower bound guess, see link) = $77.91\ \mathrm{N}$.
The snow will fall off if $gp > fr$, and this occurs at an angle of $\arcsin{\dfrac{fr}{gp}} = 32^{\circ}$.
Recalculating for the high bound friction coefficient ($1.76$), results in the snow not even falling off if the surface were vertical.
I think you already pointed out the biggest problem: any estimate is heavily dependent on the contact surface, type of snow and other conditions (temperature, wind, etc.). I haven't done a thorough looking around, but from the friction coefficients listed in this one article, I can only guess that you'll have a wide variety of ranges, making a rule of thumb difficult.