Starting from your idea: By inclusion-exclusion, for $n>3$ $\tag{1}T_n = X_n +3T_{n-1}-3T_{n-2}+T_{n-3},$ where $X_n$ is the number of "full-size" hexagons, i.e. those that touch all three sides. A full-size hexagon is determined by three positive integers $a,b,c$ (the sizes of the small triangles chopped off) with $a+b.
How many such triples ($a,b,c)$ with $\max\{a+b,a+c,b+c\} are there for each $n$? We shall assume $n\ge3$. There are $n-1\choose 2$ possibilities to choose $x,y\ge 1$ such that $x+y (think of inserting two separators into the $n-1$ gaps between $n$ pebbles, thus gouping them into $x\ge1$, $y\ge1$ and $n-x-y\ge1$ pebbles). Therefore, there are $n-1\choose2$ triples each of the forms $(1,b,c)$ and $(a,1,c)$and $(a,b,1)$. We have to subtract the $n-2$ triples each of the forms $(a,1,1)$, $(1,b,1)$, $(1,1,c)$. Then we have to add back the single triple $(1,1,1)$. What remains are triples $(a,b,c)$ with $a,b,c\ge2$. By mapping these to $(a-1,b-1,c-1)$ we biject them with the $X_{n-3}$ triples $(x,y,z)$ with $\max\{x+y,x+z.y+z\}. Therefore $\tag{2}X_n = X_{n-2}+3{n-1\choose 2}-3(n-2)+1=X_{n-2}+3{n-2\choose 2}+1 $ This would be the correct count of what corresponds to your (4) and (5). Without going into details, we see that $X_n$ is cubic in $n$, not linear as in your count; so that's where your error comes from.
We can eliminate $X_n$ by combinig equation $(1)$ with $n$ and $n-2$: $T_n-T_{n-2}=X_n-X_{n-2}+3(T_{n-1}-T_{n-3})-3(T_{n-2}-T_{n-4})+T_{n-3}-T_{n-5},$ $\tag{3}T_n = 3T_{n-1}-2T_{n-2}-2T_{n-3}+3T_{n-4}-T_{n-5}+3{n-2\choose 2}+1.$ Using this recursion and $T_{-2}=\ldots=T_2=0$ as starting values, one finds $ T_3=1\\ T_4=7\\ T_5=29\\ T_6=90\\ T_7=232\\ T_8=524\\ T_9=1072\\ \vdots $
The polynomial equation $x^5=3x^4-2x^3-2x^2+3x+1$ has $+1$ as quadruple root and $-1$ as a simple root. This suggests that there is an explicit formula $T_n=a_0+a_1n+a_2n^2+a_3n^3+b(-1)^n$. However, the cubic inhomogenuous part increases the degree so that we shall find an explicit formula $\tag{4}T_n=a_0+a_1n+a_2n^2+a_3n^3+a_4n^4+a_5n^5+a_6n^6+b(-1)^n.$ The coefficients can be obtained by plugging $(4)$ into $(3)$ and the inital values. It turns out that $ T_n = \frac{n^6}{480}-\frac{n^4}{192}-\frac{n^2}{80}+\frac1{128}-\frac{(-1)^n}{128}=\frac{4n^6 - 10n^4 - 24n^2 + 15-15\cdot(-1)^n}{1920}.$