If each LED is 3.4V and you connect 35 in series, you end up with a nominal voltage drop due to LED threshold of 119Vdc.
If you rectify full wave (Graetz bridge) connected to single-phase 120Vac you have ideally Vdcpk=170Vdc peak, and average voltage Vdcav=0.637Vdcpk=108Vdc.
The ripple is reduced using the tank capacitor you mention, and the average dc voltage is increased: you can calculate it more accurately, but considering that the loading is quite small, you can assume that the ripple will be small and the average output will be close to the peak, having subtracted the voltage drops in the diodes. Let us say VdcavC=95%*Vdcpk=160Vdc.
Now you have a dc source of 160Vdc to connect to your LED strip (35 items), with many strips in parallel. You want to control the current for, first, not to blow the LEDs, and, second, to balance the luminosity. You can add a series resistor for each strip (R1, R2, ... Rn), each one limiting the current for that strip. The current is approx 40mA, the voltage difference at nominal values is (160-119)=41V, so you end up with R=1 kohm. All resistors equal.
Now to balance luminosity you can put a second resistor in series (a trimmer) of about 10% value, so 100 ohm (or maybe 200 ohm) and you can balance each strip.
Power dissipation in the resistors is RI^2, so not an issue if you use a power resistor: 5W is advisable, 10W is better. 1 kohm is a standard value, so there is plenty of choice. For the trimmer, you will dissipate 160mW with the 100 ohm model, so better not choose the 200 ohm model, that will end up too hot for sure.
You speak of 700 LEDs, so that 35 set is replicated 20 times. Please, note that you cannot use 70 LED in series, because there is not enough feeding voltage. If your socket were 230 Vac, that would be possible.
Overall consumption is 40mA x 20 = 800 mA.