lowellbert,
Let’s try a little math, shall we?
200 mA * 200 hours = 40,000 mA-hours = 40 Ah
Now, if 3.7 equalled 5, you could do this directly with a single LiPo cell, as long as it had that capacity.
However, 3.7 does not, in my experience, often equal 5. So, you’ll need to boost the voltage, using a device such as [this. That will produce the 5 V you need. 1 joule = 1 volt-ampere-second, so:
1 joule = 3.7 V * Ain amperes * 1 second
Ain [amperes] = 1 joule/(3.7 V * 1 second)
1 joule = 5 V * Aout amperes * 1 second
Combining gives us:
Ain [amperes] = (5 V * Aout amperes * 1 second)/(3.7 V * 1 second) = (5/3.7) Aout
So, to get 40 Ah out at 5 V, you need (5/3.7) * 40 Ah ≈ 54 Ah in at 3.7 V.
But wait, there’s “less” 
: The boost circuit works at approximately 90% efficiency. So, for each joule you put in at cell voltage, you get 0.9 joules out. Correcting for efficiency, you appear to need:
40 Ah * (5/3.7) /0.9 ≈ 60 Ah
The [largest LiPo I found in quick look at the SparkFun site is 6 Ah, so you’d need ~10 of them. (That’s about 11.1 kilograms and ~$360. Here’s a shred of good news 
; SparkFun will sell you the tenth one for five cents.)
I have a few suggestions for you:
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Sharpen your metaphorical pencil and get a better estimate of the energy you need. To do that, you’ll need to estimate the fraction of the time each of the (8 rows) *(8 columns) = 64 LEDs of each color are going to be on (and at what “brightness”) and find data on the current consumption of the board’s control circuit.
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Investigate whether your energy requirements can be reduced by using “sleep” or similar functions in the Arduino and the LED controller. Better yet, if you don’t need to display all of the time, can you power the display controller down? If you don’t need to monitor for wind constantly, can you power the sensor down?
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Consider using a “bare” LED matrix, that is, one without a controller built into it. That does put more load on the Arduino, but does that matter? This approach would save you the power to run the second microcontroller and might let you use the battery power more efficiently for the LEDs.
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Consider your color choices carefully. Different colors require different amounts of electrical power to achieve the same apparent brightness.
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Unless you’re planning on having this fly (or weight is critical for some other reason), consider using a 6 V sealed lead acid (SLA) battery, such as [this. Using a 6 V battery would also save you having to boost the voltage.
Good Luck,
Eric
](http://www.power-sonic.com/images/powersonic/sla_batteries/ps_psg_series/6volt/PS-6580_11_Feb_21.pdf)](http://www.sparkfun.com/products/8484)](http://www.sparkfun.com/products/10300)