I count 238 possible row-colorings:
β¬β¬β¬β¬β¬ x1
β¬β¬β¬β¬π¨ x5
β¬β¬β¬β¬π© x5
β¬β¬β¬π¨π¨ x10
β¬β¬β¬π¨π© x20
β¬β¬β¬π©π© x10
β¬β¬π¨π¨π¨ x10
β¬β¬π¨π¨π© x30
β¬β¬π¨π©π© x30
β¬β¬π©π©π© x10
β¬π¨π¨π¨π¨ x5
β¬π¨π¨π¨π© x20
β¬π¨π¨π©π© x30
β¬π¨π©π©π© x20
β¬π©π©π©π© x5
π¨π¨π¨π¨π¨ x1
π¨π¨π¨π¨π© x5
π¨π¨π¨π©π© x10
π¨π¨π©π©π© x10
π©π©π©π©π© x1
(Okay, a much easier way to see this is: 35 = 243, minus the 5 variations on π¨π©π©π©π© which are impossible, makes 238.)
So a naΓ―ve upper bound on the number of images is 2386 β 181 trillion; and a slightly less naΓ―ve upper bound is 1 + 237 + 2372 + 2373 + 2374 + 2375 + 2376 β 178 trillion (because as soon as you see π©π©π©π©π©, the game is over).
However, I think the real answer is probably much less than that. There are a lot of images that are impossible to attain due to vagaries of the word list. For example, these two rows can never appear in the same image:
π©π©π¨π©β¬
π¨π©π¨π¨π©
There's nothing logically wrong with such a combination; e.g. if the target is ABCDE, then guessing ABEDX CBDAE will produce those colorings. But it happens that actually there are no words on the Wordle target/dictionary lists that satisfy that particular three-way cryptogram. So that knocks down our naΓ―ve estimate by roughly 6Γ5Γ2384 β 96.2 billion right there. A billion out of hundreds of trillions is just a drop in the bucket, but there are lots of drops, and they'll add up.
A naΓ―ve lower bound is 1 + 191 + 1912 + 1913 + 1914 + 1915 + 1916 β 48.8 trillion, since the target word PARSE permits 192 different row-colorings.
Another 1 + 2Γ187 + 3Γ1872 + 4Γ1873 + 5Γ1874 + 6Γ1875 β 1.38 trillion images include the row π¨π¨π¨π©π¨ somewhere, which isn't possible if the target word is PARSE, but is possible if the target word is SAINT (which permits 189 different row-colorings). We could keep chipping away, but it becomes difficult to avoid double-counting any images.
So my answer is "Somewhere between 50.18 trillion and 178 trillion."
I'm using a computer to iterate over all the possible combinations of rows, of which there are 237 + (237 choose 2) + (237 choose 3) + (237 choose 4) + (237 choose 5) + (237 choose 6) β 237 billion, and tell me how many combinations are impossible.
62 combinations of 2 rows are impossible
23507 combinations of 3 rows are impossible
3732155 combinations of 4 rows are impossible
? combinations of 5 rows are impossible
? combinations of 6 rows are impossible
So the number of impossible pictures is at least 62Γ114 + 23569Γ732 + 3732155Γ1824 + ?Γ1920 + ?Γ720... not that that helps narrow it down much.