If the IVs for both encrypted layers are transmitted in the clear, then there's a generic known-plaintext key recovery attack that works on (essentially) any encryption mode and allows the two keys to be recovered using about $3 \lceil k \mathbin/ b \rceil \cdot 2^k$ block en/decryptions, $2k \cdot 2^k$ bits of memory and $O(k \cdot 2^k)$ non-cipher computation time on average, where $k$ and $b$ are the key and block sizes of the cipher in bits (i.e. $k = b = 128$ for AES-128 or $k = 256$, $b = 128$ for AES-256). It simply goes like this:
Decrypt the first $\lceil k \mathbin/ b \rceil$ blocks of the ciphertext with each of the $2^k$ possible outer keys and the known outer IV. For each key, take the first $k$ bits of the resulting plaintext (if $k$ isn't already a multiple of $b$, which it always is for AES), append the key, and store the combined $2k$-bit string in a list.
Sort the list. (This takes $O(n \log n)$ time for an $n$-element list, where $n = 2^k$ in this case.)
Attempt to encrypt the first $\lceil k \mathbin/ b \rceil$ blocks of the known plaintext using each of the $2^k$ possible inner keys and the known inner IV. For each key, look up the first $k$ bits of the resulting plaintext in the list (using a binary search, which takes $O(\log n)$ time) to get a list of possible outer keys this inner key could match. (The average number of candidate matches you'll find per inner key is one, although of course some keys will have no matches and some will have several.) If there are any matches, encrypt one more block of the known plaintext with the inner key and decrypt one more block of the ciphertext with each matching outer key to see if they still match. If they do (which you'd expect to happen after testing half of the inner keys on average), you almost certainly have the right key pair.
A generic brute force key-recovery attack on a single layer encryption, of course, takes about $\frac12 \lceil k \mathbin/ b \rceil \cdot 2^k$ block en/decryptions on average. Thus (at least if we ignore the memory costs and only count block cipher operations) double encryption with known IVs is only (at most) $6$ times harder to break on average than single encryption, for (only!) about $\log_2 6 ≈ 2.6$ extra bits of security. Which, just to be clear, is absolutely negligible.
(Admittedly, the memory cost of this attack is not necessarily negligible, but there are also various time–memory tradeoffs that can be made to reduce it.)
Note that the generic meet-in-the-middle attack described above depends on the inner IV being available to the attacker in plain. If it's not, but is instead also encrypted with the outer key, then the attack won't work without modifications.
Of course, a trivial solution would be to simply brute force all possible values of the 128-bit inner IV in step 2. For double AES-256, this will allow the double encryption to be broken in about $2^{128} \times 2^{256} = 2^{384}$ AES encryptions, which is still less than the $2^{512}$ encryptions one would expect from the key length.
