Charge on capacitors in series  'When there are 3 capacitors connected in series, charge across each capacitor is the same.' 
Okay so say we consider three caps made of plates of the same conducting materials and same dielectric materials. 
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                      10mF   1mF

So when a charge is accumulated on the left plate of the first capacitor, by induction, this charge leaves the capacitor through the other plate. However, the second capacitor has a smaller capacitance value and hence less ability to store the charge provided by the first cap? So all the charge wouldn't flow through the second cap. So if it didn't what would happen to the excess charge?
 A: You are right that the "second (smaller) capacitor has a reduced ability to store charge" but implicit in this statement is that the voltage across each capacitor is held constant. For example, the capacitance $C$, charge $Q$, and voltage drop across the capacitor $V$ are related by $Q=CV$.  If $V$ is constant, larger $C$ means larger $Q$.
When the capacitors are in series this is not that case. The charge in the wire between the first and second capacitors must remain in that segment (the electrons can only move through a conductor and the gaps are dielectrics). This means that  $+Q$ is on one capacitor (one side of the wire segment) then $-Q$ must be on the other capacitor (the other side of the wire). Another way to put is that there must be charge conservation on an isolated conductor. 
If $Q$ is the same on both conductors (the sign isn't important), then $V_1 C_1 = Q= V_2 C_2$. So a smaller capacitor will have a larger voltage drop across it. 
Remember each point of a perfect conductor must have the same potential or it will drive currents to balance out any potential differences. At the same time, the net charge on an isolated conductor is constant due to charge conservation.
Side note 
At a certain point, if the voltage gets too large, then the dielectric could breakdown and allow a discharge.  If you put overcharge a capacitor (apply a large voltage for long enough) then the gap voltage could get too large.
A: The battery in a closed circuit does not create any charge; it "pumps" charge around the circuit.  Because of this, the negative charge on a negative capacitor plate had to be pumped there from the positive capacitor plate.  This means that for only one capacitor in the circuit, assuming that the capacitor started with no charge on either plate, the final charges on the capacitor plates are equal and opposite (1 plate is positive and 1 plate is negative) because charge must be conserved. The exact same scenario applies for capacitors in series, assuming that all capacitors started with no charge on them.  Because the negative charges on all capacitor plates had to come from positive capacitor plates, and because all capacitors are in series, the same amount of charge has to exist on all capacitor plates regardless of the individual capacitances, because charge must be conserved (i.e., the electrons on the negative plates had to come from somewhere).
