To address the original question of "Is Nikola Tesla's free energy discovery...", Tesla never created a "free energy device". One of his noted ideas, however, was a system to intentionally transmit power wirelessly. Power companies don't intentionally radiate energy (as it's a pure loss for them).
As an aside, Nikola Tesla was one of the first true electrical engineers, taking arcane, hard-to-understand forces and turning them into marketable solutions. While there is no doubt he was brilliant, this revolutionary engineer would quickly tell you that if you wanted to harvest naturally occurring electrical fields (not those he intentionally radiated) it would take an antenna (or an array of them) on a truly grand scale.
Regarding the document you linked:
Chapter 4 - Tesla's Radiant Energy Device
This chapter discusses a patent by Tesla which discusses using either the photoelectric effect via "ultra-violet light [...] and Roentgen rays [X-rays]" to generate a positive charge by ejecting electrons, or cathodic rays to capture electrons and generate a negative charge.
While you might be able to use the photoelectric effect from solar UV on metals, with great care, you are going to derive an extraordinary small current, certainly far less than you would get with a photovoltaic (solar) cell. PV cells use the photoelectric effect, but within a semiconductor.
Chapter 5 - The Tesla Coil
Tesla coils are essentially antennas that can radiate and receive a great deal of power. In order to actually capture an appreciable amount, much, much more must be broadcast on the particular wavelength that the coil is tuned to. Because they are tuned, they cannot capture broadband noise
The typical rule of thumb for these strips is injecting power every 5m or better. Power should be injected at both ends, for the best color matching. FPC, flexible printed copper has relatively high resistance and it quickly adds up.
You need 1890W ( (15m * 7) * 3.6A ) * 5v. You mentioned the supply's power efficiency at 80%, but that should already be accounted for in it's stated output capacity. On the other hand, a 20% margin of safety, as to not drive the supplies at their maximum output, is a good idea. 2300W it is.
You have two options. Get a few 5v high current supplies, and run some cables in parallel to each 4~5m section. Due to the high current, this can get pretty expensive, as each 5m section takes 18 Amps. A quick calculation with a AWG calculator, At a 32 foot run (5 Meters to and from, you have to double the cable run) with 18A at 6V, you get a 0.6V drop... USING 10 AWG. That's 10% wasted in pretty thick and expensive cable. You could go with a 12v supply instead, at 18A, 10M, 20AWG, would result in 50% voltage drop, giving you 6v at the target end. Still not efficient.
The other option, is to get some high voltage, low current supplies (Say 48v), and some small dc-dc buck switchers able to handle 18A output. You only need 90W at the target. At 48V, 90 Watts is a little under 2 amps. 2.5 Amps, to adjust for the cable drop and switcher efficiency. This way you only need some 22AWG cable, resulting in a 2.5% voltage drop, which doesn't effect the switcher at all. You could even run it all in series, as the resistance drop at 22 AWG and 48v can be very negligible with the right switchers. And you only need a 48v 50A (2400W) supply to power it all. Way better than 5v 480A and high gauge cables.
Best Answer
The power rating of a device is the maximum power and is usually given in watts for real power and volt-amperes for apparent power for AC systems. Also not uncommon is the indication of active power and power factor.
DC power will always be a real power value, as there's no such thing as real/reactive power in a DC circuit. (That could be the case with above mentioned LED device, if DC power ratings are given.)