Actual Question below.
Consider the use of double encryption applied to the AES algorithm with two 128-bit keys. How much storage and computation would be required to execute a meet-in-the-middle attack?
Double encryption applied to the AES-128 algorithm and two random keys is theoretically breakable from 3 plaintext/ciphertext pairs by the generic Meet-in-the-Middle attack with 2128 encryptions for building a dictionary, and an expected 2127 attempts each requiring a about one decryption and two encryptions (the later is for checking an average of one candidate key pair). This totals to an expected 5⋅2127 AES-128 operations. That's so large as to be currently impossible: the realm of feasibility nowadays stops somewhere a 275 to 2100, with the later acknowledged good to 2020 by many authorities. Thus we have a margin like a thousand millions, which I guess is fine for a good two decades, save for fairies or imminent breakthrough in quantum computers making them useful for such tasks.
The amount of memory required for the schoolbook MitM attack is 2128 words each 128-bit, that is 2135 bits. Nowadays, a silicon wafer is seldom less than
80µm 50µm thick (these are used in Smart Cards, and obtained by milling normal wafers rather than by using ultra-thin wafers), a DRAM cell occupies an area of at least 4F2 with feature size F=10nm, thus we are talking 0.9⋅1021m3 of silicon; that's 80% of the volume of earth, which contains only 15% silicon. Quick, make these wafers even thinner or (more realistically) start making 3D silicon that is not stacked wafers!
There is a well-known refinement of MitM that requires a feasible amount RAM, but more computation than the generic MitM. See Paul C. van Oorschot and Michael J. Wiener: Parallel Collision Search with Cryptanalytic Applications (in Journal of Cryptology, 1999).
Interestingly, it is not known any generic technique that uses moderate amount of RAM and breaks two-key triple encryption of some b-bit block cipher with cost commensurate to 2b or even 23 b /2 encryptions.
fgrieu answered this in a footnote elsewhere:
Double AES with two 128 bit keys would be significantly slower than single AES with a 256 bit key (20 rounds versus 14). Also double AES-128 would theoretically be vulnerable to a Meet-in-the-Middle attack, when AES-256 is not.
So how much storage and computation a meet-in-the-middle attack would take doesn't matter as there is no real reason to be using double AES-128 (as it's slower and probably weaker than AES-256).