Generating random words by grammar
Your process is a textbook example of a branching process. Starting with one $E$, we have an expected $3/2$ many $F$s, $9/4$ many $T$s, and so $9/8$ many remaining $E$s in expectation. Since $9/8 &gt; 1$, it is not surprising that your process often failed to terminate.

To gain more information, we need to know the exact distribution of the number of $E$-offsprings, which is given by the following generating function (see the Wikipedia article linked to above):
$$
h(z) = \frac{33}{128} + \frac{7}{16}z+ \frac{15}{64}z^2 + \frac{1}{16}z^3 + \frac{1}{128}z^4.
$$
This implies that the extinction probability is $d \approx 0.717778143742483$ (by solving $h(z) = z$). This is an upper bound on the probability that your procedure terminates.

We can easily recover your numbers by considering iterates of $h$. The probability that the process terminates in $3k$ steps is $h^{(k)}(0)$. So computing $h(0),h(h(0))-h(0),h(h(h(0)))-h(h(0))$ and so on, we recover the figures in your table.

When $k$ is large, $h^{(k)}(0) \approx d$. We can compute $h'(d) \approx 0.882115879977412$. We have
$$
\frac{d - h^{(k)}(0)}{d - h^{(k-1)}(0)} =
\frac{h(d) - h(h^{(k-1)}(0))}{d - h^{(k-1)}(0)} \approx h'(d).
$$
It follows that
$$
d - h^{(k)}(0) \propto h'(d)^k.
$$
Therefore the probability that the process terminates in exactly $3k$ steps is
$$
h^{(k)}(0) - h^{(k-1)}(0) =
[d - h^{(k-1)}(0)] - [d - h^{(k)}(0)] \propto h'(d)^{k-1} - h'(d)^k \propto h'(d)^k.
$$
Empirically, we can check that the constant of proportionality is roughly $0.0248011196615094$.