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Is it (2^48) possible addresses X 48 bits = 1536 terabytes ??

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Why would you want a list of all the numbers from 0 to 2^48 (281 474 976 710 656)? – martineau Feb 20 '11 at 8:41
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There are 248 (281474976710656) possible EUI-48 addresses, which would occupy 248 × 48 bits or 1688849860263936 eight-bit bytes, which is 1536 terabytes or 1.5 petabytes.

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That sounds about right.

Plus some Bytes for the filesystem.

And a lot more bytes if you want to store everything as textfile. In this case we are talking about 2^48 (address range) * 48 bit (per address) * 8 bit (per char, assuming you store everything in ASCII) = about 12288 terabytes. And don't forget dividers between the single addresses - if required.

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If you're going to store in ASCII, you wouldn't store '0' or '1' for everything, you'd use hex. With 8-bit chars, you cut the expansion from 8:1 to 8:4, or 2. 48-bit address is 12 hex digits; 96 bits with 8bit characters or 84 with seven-bit ASCII. You could put two 48-bit addresses into 21 bytes of storage, even with character encoding. – mpez0 Feb 20 '11 at 1:28
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