At what acceleration will electron flow be disturbed?
First, let's assume a horizontal plate capacitor with a free electron in its field. To cancel out gravity (i.e. let the electron hover), you need an electric field given by
$$Ee=m_eg$$
$$E=\frac{m_eg}{e}=\frac{9.1\cdot10^{-31}kg\cdot 9.81m/s^2}{1.6\cdot10^{-19}C}=55.8\cdot10^{-12}V/m$$

This is a really really tiny field, and it would be impossible to measure it, as your measurement devices will be affected by gravity, too.

However, there were experiments to measure effects of forces coupling to the mass of electrons:

K. V. Nichols used a spinning disc of metal and tried to measure the voltage between the axis and the outer edge:


Source (Google Books): Richard Becker: Theorie der Elektrizität Band 1

The expected voltage is 

$$U=\frac{m_e\omega^2R^2}{2e}$$

Nichols used a radius of \$R=0.1m\$ and \$\omega=100/s\$, which leads to \$U=3\cdot 10^{-10}V\$. This again is an incredibly low voltage, and Nichols also failed to measure it. Other effects at the contact surfaces were larger. The problem is that these effects increase with speed / radius of the disc, so there's no way to increase the desired effect by increasing speed or radius.

Richard C. Tolman used an other setup, consisting of a coil spinning around its axis. When accelerating / decelerating the coil, there should be a voltage between the ends of the wire due to the inertia of electrons:


Source as above

The expected voltage is 

$$U=\frac{m_el}{e}\frac{dv}{dt}$$

\$l=\$ is the length of the wire.
The physics book with these pictures doesn't give dimensions, but if we assume a wire of 100m length and an acceleration of 1g, this gives \$5.5\cdot 10^{-9}V\$. This is about 20 times the value from the other experiment, but you can easily achieve a higher acceleraton of 10g and more.



Finally, you can re-interpret the last formula. It gives the voltage inside a piece of metal over a distance l along which a constant acceleration of a=dv/dt exists. 
But it also shows that the effect is negligible. Even for Burj Arab, that 830m building, there will be a voltage of \$U=4.6\cdot10^{-8}V\$ across wires from ground to the top due to gravity.
And as said, if you connect a multimeter, gravity affects its cables, too, and you would not measure any voltage.

(Yes, physicist here...)