I2 is the charge pump current through pin No.2 which is given on the datasheet as about 165uA at Vcc of 7.5V.
For the circuit you're considering, matching the output voltage to the display input sensitivity at the maximum frequency you want to display is more important than the max possible input frequency. What that means is this:
The full-scale input of each LM3914 is 1.25V. Since two LM3914s are cascaded here, the input for full-scale display is 2.5V. Therefore, if you want the display to show full scale at 250Hz, the LM2917's output should be 2.5V at 250Hz. Taking the formula given on the datasheet -
Vo = Vcc*fin*C1*R1 or, if we use the part Nos. in the circuit you uploaded: Vo = Vcc*fin*C6*R5
This can be arranged as C6*R5 = Vo/(Vcc*Fin)
Since Vcc = 7.5V, fin is 250Hz max, Vo = 2.5V at 250Hz
Then C6*R5 = 2.5/1875 = 0.00133
If we use the value of C6 already given as 47nF or 0.047uF
Then R5 = 0.00133/0.047 = 0.0283. This calculated value of R5 is in Megohms since C6 was given in uF
Therefore R5 = 28.3k which is within the adjustment range for the 50k pot
In practice, it is best to put a fixed resistor of 10-15k in series with the 50k pot.
Using the values of I2 = 165uA and Vcc = 7.5V
fmax = 468 Hz. This result assures us that the IC will still be operating in the linear region at 250Hz.
But the output voltage Vo at fmax is about 4.68V which is well above the full-scale input of the display section. We've already calculated the component values to give full-scale display at 250Hz. Therefore any input frequency above 250Hz will make no difference in the display.