Why is the voltage across the capacitor unbound?
To understand what is going on here, let's first make sure we understand two things in a well-defined way:

A capacitor stores charge on its plates. As it does, this difference in charge forms a potential difference (a voltage) between the plates. The ratio between the amount of charge on the plates (or in other words, in the capacitor) is given by the capacitance:

$$ C = \frac{Q}{V}$$

Where \$C\$ is the capacitance in [F] Farad, \$Q\$ is the charge in [C] Coulomb and \$V\$ is the voltage in [V] Volt.

We will also need to understand Current. Current is the (net) amount of charge that passes a certain cross section of the wire in a given amount of time. In other words, the current describes the flowrate of charge. In equations:

$$ I = \frac{Q}{t} $$

Where \$I\$ is the current in [A] Ampere (not amp-age or amperage! Those are historical terms), \$Q\$ is again the charge in Coulomb, and \$t\$ is the time in [s] seconds.

Now, with all this math behind us, what is going on here? 

A current source will output a certain amount of current. This means that it is "pushing" a certain amount of charge out of it's "exit" terminal (the one the arrow points to) and sucking the same amount of charge into it's "entrance" terminal. This current flows into our capacitor. In other words, a constant amount of charge is continuously being pushed into the capacitor. 
If we write the charge in the capacitor as a function of time, we can see from our definition of current:
$$Q(t) = I \cdot t $$
Let's plug that into the voltage equation from the definition of the capacitor:
$$V(t) = \frac{Q(t)}{C} = \frac{I\cdot t}{C} = \frac{I}{C}\cdot t$$
In other words, the voltage on the capacitor is a function of time, increasing linearly with time! This is what you are observing: as time progresses, more and more charge is being pushed into the capacitor, creating more and more potential difference.

How much energy is stored? We know that energy is work (=power) over time, or:
$$E = P \cdot t = V_{\text{avg}} \cdot I \cdot t$$

The trick here is that the voltage is not constant. We can solve this by filling in our equation for voltage we found earlier. Not that since the voltage is ramping up linearly, the average voltage is just half of the current voltage:

$$V_{\text{avg}} = \frac{V(t)}{2} = \frac{1}{2} \cdot \frac{I\cdot t}{C}$$

$$E = V_{\text{avg}} \cdot I \cdot t =  \frac{I\cdot t}{2C} \cdot I \cdot t = \frac{I^2 \cdot t^2}{2C}$$

NOTE: This is all assuming the current is constant. If it is not, we have to resort to an integral, and we can write that 

$$Q(t) = \int_{t_{\text{start}}}^{t_{\text{end}}} I(t) dt$$ 
and 
$$V(t) = \frac{Q(t)}{C} = \frac{\int_{t_{\text{start}}}^{t_{\text{end}}} I(t) dt}{C}$$

The energy can now be calculated with 

$$E = \int_{t_{\text{start}}}^{t_{\text{end}}} V(t) I(t) dt$$

Also note that in practice this will not be possible with "regular" components, as the voltages required will pretty much destroy any component hobbyists have access to.