Name the faces of the $3\times3$ cube touching the start A, B, and C. Name the faces touching the finish vertex D, E, and F. Any valid path lies within just two of these faces - one of A, B, and C, and an adjacent one of D, E, and F.
The number of paths that lie entirely within faces A and D is $9\choose3$. (Unfold the two faces into a $3\times6$ rectangle.) Similarly, there are $9\choose3$ paths that lie entirely within faces A and E, if A and E are adjacent, etc. There are 3 ways to choose one of A, B, and C and for each choice, two of the faces D, E, and F are adjacent, so this method counts $6{9\choose3}$ paths.
Unfortunately, this counts paths twice if they begin or end (but not both) by traversing an entire edge of the larger cube, and it counts paths three times if they both begin and end by traversing an entire edge of the larger cube.
The number of paths counted exactly twice is $3\cdot({6\choose3}-2)+({6\choose3}-2)\cdot3=108$, and the number of paths counted exactly three times is 6.
All told, then, there are $6{9\choose3} - 108 - 2\cdot6 = 384$ paths.
I think your idea is basically right. Remember, however, that there are two types of paths that have been double-counted: paths that pass through a corner adjacent to the initial corner, and paths that pass through a corner adjacent to the final corner. Therefore you need a factor of $6$ rather than $3$ in your second term.
Here's an inclusion-exclusion argument that clarifies things—at least for me: let the ant start at the front, bottom, left corner, and end at the back, top, right corner. Let the faces adjacent to the initial corner be called $F$ (front), $B$ (bottom), and $L$ (left), and let the faces adjacent to the final corner be called $F'$ (back), $B'$ (top), and $L'$ (right).
As you say, the ant's path lies within two faces. One of these will be an unprimed face; one will be a primed face. Define $P(X,X')$ to be the set of paths that lie within faces $X$ and $X'.$ We have
$$
\lvert P(X,X')\rvert=\begin{cases}0 & \text{if $X$ and $X'$ are opposite each other,}\\
\binom{9}{3} & \text{otherwise.}\end{cases}
$$
To use an inclusion-exclusion argument, we need to know the sizes of intersections of sets $P(X,X').$ As stated in your solution, intersections like $P(B,F')\cap P(L,F')$ contain $\binom{6}{3}$ paths since such paths must traverse the edge where $B$ and $L$ meet, leaving $\binom{6}{3}$ ways to get from one corner of $F'$ to the other. The same is true of intersections like $P(F,B')\cap P(F,L').$ There are six such intersections in all. The other non-empty two-way intersections are sets like $P(F,B')\cap P(B,L'),$ which contains the single path that traverses the edge where $F$ and $B$ meet, the edge where $F$ and $L'$ meet, and the edge where $B'$ and $L'$ meet. There are six of these as well. Finally, the only non-empty three-way intersections are sets like $P(F,B')\cap P(F,L')\cap P(B,L'),$ which contains the single path described above. There are six of these too.
Inclusion-exclusion then says that the number of paths is the sum of the sizes of all the sets $P(X,X')$ minus the sum of the sizes of all two-way intersections plus the sum of the sizes of all three-way intersections. This yields
$$
6\cdot\binom{9}{3}-\left(6\cdot\binom{6}{3}+6\cdot1\right)+6\cdot1=6\cdot\binom{9}{3}-6\cdot\binom{6}{3}=384.
$$
Best Answer
As another way to see it, note that we can describe any path along the cube as a string of $X$'s, $Y$'s, and $Z$'s, where X means "change your $x$-coordinate from 0 to 1 or vice versa," and similarly for $Y$ and $Z$. What's more, the string corresponding to any closed path along the edges of the cube will have an even number of $X$'s, $Y$'s, and $Z$'s.
So the cases are:
Strings with 6 identical symbols. There are three such strings: $XXXXXX$, $YYYYYY$, and $ZZZZZZ$.
Strings with four symbols of one type and two of another. For a given choice of the two types of symbol ($X$ and $Y$ respectively, say) there are $\frac{6!}{4! 2!} = 15$ such strings. There are further 6 possible choices for the types of symbols in the string (noting that, for example, 4 $X$'s and 2 $Y$'s is distinct from 4 $Y$'s and 2 $X$'s.) So we have $15 \times 6 = 90$ such strings.
Strings with two symbols of each type. There are $\frac{6!}{2!2!2!}=90$ such strings.
So all in all there are $3 + 90 + 90 = 183$ such strings and therefore 183 such paths on the cube.