How to determine forward voltage for overdriven LEDs
We can model our LED as a resistor in series with a diode.

$$V_{LED} \approx V_R+V_D$$

For a diode

$$I=I_\mathrm{S} \left( e^\frac{V_\text{D}}{n V_\text{T}} - 1 \right)$$

$$\frac{I}{I_\mathrm{S}} = e^\frac{V_\text{D}}{n V_\text{T}}-1 $$

Asssuming \$V_D &gt;&gt; n V_\text{T}\$

$$\ln\frac{I}{I_\mathrm{S}}\approx  \frac{V_\text{D}}{n V_\text{T}}$$

$$V_\text{D} \approx n V_\text{T} \ln\frac{I}{I_\mathrm{S}} = n V_\text{T} \ln I - n V_\text{T} \ln I_\mathrm{S}$$

For a resistor 

$$V_R = IR$$

Combining

$$ V_{LED} \approx n V_\text{T} \ln I - n V_\text{T} \ln I_\mathrm{S} + IR$$

The middle term is a constant (assuming constant temperate).

For small currents the first term dominantes the resposense, but for large currents the last term dominates the response, the graph from your data sheet clearly shows the last term dominating, that is the graph is roughly linear over the range shown.

A very approximate reading of your graph gives us a linear equation.

$$I = 40 (V - 3)$$

So at 12V I would expect "only" 360ma.