Prove or disprove that $2^n$ divides $T_{2^n}$ for $n > 2$.

I'll show that $2^{n+1}\mid\mid T_{2^n}$ for all $n\ge 5$. Together with casework for $n<5$ this gives a proof of your claim.

Lemma: For all integers $n,m\ge 1$ we have $$T_{n+m}=T_{m}T_{n-1}+T_{m-1}T_{n}+T_{m}T_{n}+T_{m+1}T_{n+1}.$$ Proof: Straightforward induction on $m$. $\square$

Proposition: For all integers $n\ge 5$: $$T_{2^n}\equiv 2^{n+1}\pmod{2^{n+2}}.$$ Proof: We shall use induction on $n$ to prove simultaneously the following congruences for $n\ge 5$: $$\begin{cases} T_{2^{n}-2}&\equiv -1&&\pmod{2^{n+2}}\\T_{2^{n}-1}&\equiv 2^{n-1}+1&&\pmod{2^{n+2}}\\T_{2^{n}}&\equiv 2^{n+1}&&\pmod{2^{n+2}}\\T_{2^{n}+1}&\equiv 5\cdot 2^{n-1}&&\pmod{2^{n+2}}\\T_{2^{n}+2}&\equiv 2^{n}+1&&\pmod{2^{n+2}}\end{cases}$$ This is true for $n=5$. Assume it to be true for some $n\ge 5$. Then by the lemma, $$\begin{align}T_{2^{n+1}}&=T_{2^n-1}T_{2^n}+T_{2^n-2}T_{2^n+1}+T_{2^n-1}T_{2^n+1}+T_{2^n}T_{2^n+2}\\&\equiv(2^{n-1}+1)2^{n+1}-5\cdot2^{n-1}+(2^{n-1}+1)\cdot 5\cdot 2^{n-1}+2^{n+1}(2^n+1)\pmod{2^{n+3}}\\&\equiv2^{n+2}\pmod{2^{n+3}}\end{align}$$ For $T_{2^{n+1}-1}$ and $T_{2^{n+1}+1}$ we use something similar, so I let that to you. Finally, $T_{2^{n+1}-2}$ and $T_{2^{n+1}+2}$ are found using the recursion formula. This completes induction. $\square$

A solution by matrix form (as suggested by Ihf): We can write:

$\left(\begin{array}{c} T_{n+1}\\ T_{n+2}\\ T_{n+3} \end{array}\right)=\left(\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 1 \end{array}\right)\left(\begin{array}{c} T_{n}\\ T_{n+1}\\ T_{n+2} \end{array}\right)$

So if $M$ is the matrix of multiplication, we can write $(T_n,T_{n+1},T_{n+2})^t=M^n\cdot(0,0,1)^t$.

In particular $T_n$ is the value in the right-top corner of the matrix $M^n$, so we have to show that for $n>2$ the righ-top corner of the matrix $M^{2^n}$ is divisible by $2^n$. Let's do it by recursion: let $n$ be an integer such that we can write the matrix $M^{2^n} \pmod{2^n}$ in the following form (for example $n=3$):

$\left(\begin{array}{ccc} 2^{n-1}+1 & 2^{n-1} & 0\\ 0 & 2^{n-1}+1 & 2^{n-1}\\ 2^{n-1} & 2^{n-1} & 1 \end{array}\right)$

Now we lift it to a matrix $\pmod{2^{n+1}}$, clearly we have more possibilities, so we will write it in the following form:

$\left(\begin{array}{ccc} 2^{n-1}+1+\delta_1\cdot2^n & 2^{n-1}+\delta_2\cdot2^n & \delta_3\cdot2^n\\ \delta_4\cdot2^n & 2^{n-1}+1+\delta_5\cdot2^n & 2^{n-1}+\delta_6\cdot2^n\\ 2^{n-1}+\delta_7\cdot2^n & 2^{n-1}+\delta_8\cdot2^n & 1+\delta_9\cdot2^n \end{array}\right)$

where $\delta_i=0,1$.

If we calculate the square of the matrix (always mod $2^{n+1}$) we found a matrix in the same form of the case $2^n$.