Let $1$, $\omega$, and $\omega^2$ be the cube roots of unity. So $\omega=e^{2\pi i/3}$. I don't like the letter $t$ in the context you are using it, so will use $n$.
By the Binomial Theorem, we have
$$(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k.$$
Now let
$$F(x)=(1+x)^n +(1+\omega x)^n+(1+\omega^2 x)^n.$$
If we expand, the coefficient of $x^k$ is
$$\binom{n}{k}(1+\omega^k+\omega^{2k}).$$
If $k$ is divisible by $3$, then $1+\omega^k+\omega^{2k}=1+1+ 1=3$. If $k$ is not divisible by $3$, then $1+\omega^k+\omega^{2k}=0$.
So
$$\frac{F(x)}{3}=1+\binom{n}{3}x^3+\binom{n}{6}x^6+\binom{n}{9}x^9+\cdots,$$
(finite sum). Put $x=1$. We get
$$\frac{F(1)}{3}=\frac{(1+1)^n+(1+\omega)^n+(1+\omega^2)^{2n}}{3}=\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\binom{n}{9}+\cdots.$$
For the probability, divide by $2^n$.
So our probability, simplified a bit, is given by
$$\frac{1}{3}+\frac{(1+\omega)^n +(1+\omega^2)^n}{3\cdot 2^n}.$$
For computation, we will want to fool around with $(1+\omega)^n+(1+\omega^2)^{n}$. Note that $1+\omega$ and $1+\omega^2$ are conjugates. Express $\omega$ in polar form $re^{i\theta}$. It will involve a $60^{\circ}$ angle. Then $(1+\omega)^n+(1+\omega^2)^n =r^n (e^{i n\theta}+e^{-i n\theta})$. That will just be $2r^n \cos(n\theta)$. Can simplify further, getting a "mod" expression, since the cosines cycle.
Another way: Here is a brief sketch of another approach. Let $P_0(n)$ be the probability that the number of tails is congruent to $0$ modulo $3$, and let $P_1(n)$ the probability that the number of tails is congruent to $1$ modulo $3$. For symmetry we should also have a $P_2(n)$, but it can be dispensed with, since it is $1-P_0(n)-P_1(n)$.
We can get recurrences for $P_0(n)$ and $P_1(n)$ in terms of each other. The first is
$$P_0(n)=\frac{1}{2}P_0(n)+\frac{1}{2}P_2(n-1)=\frac{1}{2}P_0(n-1)+\frac{1}{2}(1-P_0(n-1)-P_1(n-1)).$$
This is because in our $n$ tosses, the number of tails is congruent to $0$ mod $3$ if (i) it was congruent to $0$ and we got a head or (ii) it was congruent to $2$ and we got a tail.
Similarly, but more simply,
$$P_1(n)=\frac{1}{2}P_1(n-1)+\frac{1}{2}P_0(n-1).$$
We can solve this system of two recurrences by matrix methods. It is more common in undergraduate courses to get from our two first-order recurrences a single second-order linear recurrence for $P_0(n)$. You may have seen methods to deal with these (find roots of characteristic polynomial, $\dots$).