Power seems to be an important aspect in your design, so you might consider using different LEDs. Low-current LEDs for instance take 2mA instead of 20mA.
Or maybe 5mA through your existing LEDs will produce enough light for your taste.
Another angle would be to design your LED drive circuit for as low a voltage as you can. If you can design it for 3V and you use a switched power supply to create this 3V from 5V (or maybe 12V), you have saved 40% energy. This is especially effective when you use a battry, which produces a lower voltage over time.
I guess that when you want to iluminate (nearly) all LEDs you can get by with illuminating them each a bit less. This logic could be incorporated in your software, so the worst case current would be reduced.
You must design your circuitry for the worst case, but for battery life it might be more realistic to calculate with averaged cases. So get some more info on what you want to show when you need to know how a given battery will last.
As often, your one good question results in an avelange of questions, and some questioning of your basic assumptions. That's system design :)
Background:
I have designed a number of LED lighting products which are manufactured in China.
I have several cylindrical LED flashlights that have a large number of LEDs in them
... Are there ultra-bright LEDs that you can drive directly off of 4.5 volts without a current limiting resistor? Or are there special purpose ultra bright white LEDs made for 4.5 volt supply that have internal current limiting resistors?
No and no, unfortunately.
Many LED lights are constructed as you describe, with multiple white LEDs wired in parallel and connected essentially directly across the battery.
They are junk.
They are not "designed".
They build them this way "because they can" and they work well enough to be able to sell them.
When supplied with 4.5V + the LEDs are driven well above their maximum design rating and their lifetimes are greatly shortened. The LEDs used are typically low lifetime low cost devices.
Follow-up question: Does anybody know if the 12 volt LED bulbs that are in landscape lights have a voltage regulator in them?
The 12 volt LED strips usually use 3 LED die in series plus a series resistor.
Turn on / turn off time is liable to be sub `1 microsecond if capacitors are not used downstream of the switch.
Current is set to be "about right" at 12 Volts so will vary substantially if used in an automotive context where several volts of variation occurs. Many strips use individual LEDs but some use 3 die per package LEDs with all 3 independent die wired in series. It is possible but not certain that strips with individual LEDs will run somewhat cooler due to a lower concentration of Energy per package.
Lifetime of these LEDs may be better as the series resistor means that they are somewhat more properly driven. I have seen very substantial variations in output of similarly appearing strips. The brightness bears no obvious relationship to LED specifications and a brighter strip may simply reflect a manufacturers 'marketing decision'. You can get a range of LEDs per metre but current drain and number of LEDs are not directly related.
White LEDs are typically have a voltage drop in the 3.0 - 3.5V range at rated current.
Current increase tends to be exponential with voltage and at 4.5V almost any LED would self destruct almost instantly. The "saving grace" (if it can be called that) is that the combination of small batteries and many LEDs means that the batteries are unable to produce more than 'vastly too much' current when the batteries are new. Any light constructed in this manner demonstrates a total lack of concern and/or knowledge by the manufacturer.
Adding even a single common series resistor makes a substantial improvement in voltage/current profile and a resistor per LED would greatly assist current balancing between LEDs.
Added May 2016
Harper commented:
OP is asking about LED bulbs, not strips. Those are commonly made as screw-in replacements for incandescents. Some have a resistor, but many have a switching buck converter which will accept a range of voltages from 12-30V or higher. The LED series voltage is quite close to 12V actual, so if voltage drops much below 12V the buck converter will go to 100% duty cycle and simply pass the voltage through, causing the LEDs to dim rapidly.
My answer addressed LED strips as I noted, which the OP did not ask about, as Harper noted :-).
Harper's comments above are correct where applicable. I have not seen a bulb with a buck converter internally, but no doubt they exist. White LEDs have Vf typically in the range 2.8V - 3.5V. 2.8V is unusual and usually only seen in reasonably modern LEDs or ones operated well under full power. At 12V nominal, 4 LEDs have 12/4 = 3V each available. Allowing a small voltage drop in connectors and wiring 4 LEDs with Vf of 2.8V to 2.9V would be able to be operated at full power. In real world situations with Vin able to be somewhat below to substantially above 12V, 4 LEDs in series will often work but 3 x LEDs in series plus a series resistor is 'safer'. Bulbs may not match strips in configuration, but all 12V LED strips that I have seen use 3 LEDs in series plus a resistor.
Best Answer
Power
According to the WS2812 datasheet (I can't find a WS2812B datasheet and you didn't provide a link) the integrated LED chip and controllers run on a supply voltage of 6 to 7 V and a current of 20 mA per R, G and B LED. Total all-on power demand given by \$ P = VI \$ will be \$ 2000 \cdot 3 \cdot 6 \cdot 0.02 = 720 W \$. You have underestimated on your power budget.
Current
Max current is simply the sum of all the currents when all LEDs are on. \$ I_{MAX} = 2000 \cdot 3 \cdot 0.02 = 120 A \$.
Battery
Batteries are typically rated in Ah (ampere-hours). The manufacturer will provide an amp-hour rating at a specific discharge rate. If you discharge at a higher rate the amp-hour rating decreases.
Ignoring that for a moment, at 6 V (no voltage conversion) you need 120 A x 16 h = 1920 Ah capacity. Let's round it up to 2000 Ah. We can de-rate, for example, if we assume that on average 50% of the LEDs will be on, to 1000 Ah.
As an example, the 6 V Yuasa battery pictured above has a 12 Ah capacity. You would need \$ \frac {1000~Ah}{12~Ah} = 83 \$ to give 1000 Ah capacity. Weight: 83 x 2.05 kg = 171 kg. Guide price: 83 x €16 = €1,328 from Farnell.
Charging
Charge current for an ideal battery will be \$ \frac {Ah~capacity}{charge~time} \$. For an 8-hour charge time that would work out at \$ \frac {1000}{8} = 125~A \$.
Comments
These are high currents at low voltages. A 100 to 200 A, 6 V lead acid battery charger will be difficult to find.