Capacitors charging in a DC circuit
Time constant \$= \tau = RC\$.

\$ R = 3 \cdot 470 mF\$ (based on one of two values you give) so:

$$\tau = 10 \cdot 0.47 F \cdot 3 \approx 15 \text{ seconds}$$

That is one exponential time constant or about 65% of full. Each successive time constant fills it about another 65% of the remaining voltage. 

However, the LED (assume \$V_{operate} \approx 2V\$) has a current drain of:

$$I = \frac{V}{R} = \frac{V_{cap}-V_{led}}{R2} \approx 2 mA$$

about \$2mA\$ per volt on caps above 2 volts. So, about \$2 mA\$ at \$3V\$, \$6 mA\$ at \$5V\$, or \$10 mA\$ at \$7V\$.

Light level between \$3V\$ and \$7V\$ is about 10 mA : 2 mA = 5:1 but your eye sees it as a much lower range than that. You need a means of delinearising LED response or, easier, using some other indicator.  A bar graph voltmeter or a small analog meter would do a better job for you.

LM3914 based bar graph voltmeter
And similar