7. Given a $2008 \times 2008$ chessboard, each small square has a different color. In each small square of the chessboard, fill in one of the 4 letters $C, G, M, O$. If in every $2 \times 2$ small chessboard there are the 4 letters $C, G, M, O$, then the chessboard is called a "harmonious chessboard". How many different harmonious chessboards are there?
7. There are $12 \times 2^{2088}-24$ different harmonious chessboards.

First, prove a conclusion.
Conclusion: In each harmonious chessboard, at least one of the following situations must occur:
(1) Each row is an alternating sequence of two letters;
(2) Each column is an alternating sequence of two letters.
In fact, assume that a certain row is not alternating, then this row must contain three adjacent small squares filled with different letters.

Without loss of generality, assume these three letters are $C$, $G$, and $M$ (as shown in Figure 5). It is easy to see that $X_{2}=X_{5}=O$, and $X_{1}=X_{4}=M$ and $X_{3}=X_{6}=C$ (as shown in Figure 6).

Similarly, it can be deduced that these three columns are all alternating sequences of two letters. Thus, it is easy to conclude that each column is an alternating sequence of two letters.
Returning to the original problem.
Next, calculate the number of harmonious chessboards.
If the leftmost column is an alternating sequence of two letters (such as $C$ and $M$), it immediately follows that the columns with odd indices are all alternating sequences of these two letters, and the columns with even indices are all alternating sequences of the other two letters (such as $G$ and $O$). The first letter in each column can be either of the two letters in that column.

It is easy to verify that any such filling can result in a harmonious chessboard. Therefore, there are $\mathrm{C}_{4}^{2}=6$ different ways to choose the two letters for the first column, and there are $2^{208}$ ways to determine the first letter in each column. Thus, there are $6 \times 2^{2008}$ ways to fill the board such that each column is alternating. Similarly, there are $6 \times 2^{2008}$ ways to fill the board such that each row is alternating.

The next step is to subtract the fillings that have been counted twice—those where both rows and columns are alternating.

Clearly, any arrangement of the four different letters in the $2 \times 2$ square at the top-left corner can be extended to the entire chessboard to form a harmonious chessboard, and both rows and columns are alternating.

In fact, as long as the first two columns are filled to be alternating, and then all rows are filled to be alternating, it is sufficient. Conversely, such a double-alternating filling is uniquely determined by the $2 \times 2$ square at the top-left corner. There are $4!=24$ ways to arrange the four different letters in the $2 \times 2$ square at the top-left corner. Therefore, there are 24 different ways to fill the board such that both rows and columns are alternating, leading to the above result.