I would propose an alternative - maybe easier - approach, trying to give a formula that provides a generalized answer for the probability of three events in a row, given the probability $p $ that the event occurs, as a function of the sample size $n$. As shown below, this approach may also be useful to determine an additional issue of this problem highlihted in one of the comments, i.e. the sample size needed to "guarantee" a success with a given confidence level.
Let us call $P (n) $ the probability that our event occurs three times in a row within a sequence of $n $ trials. Also, for each trial, let us denote as Y the case that the our event occurs, and as N the case that it does not occur. So, in the specific case described in the OP, we are looking for the probability of getting a triple YYY in a sequence of $n $ elements when $p=0.7$.
As a general rule, we can note that, passing from a sample size $n $ to a sample size $n+1$, the probability that our event occurs three times in a row increases only because there is the possibility that the triple YYY, not obtained in the first $n $ trials, occurs as a final triple in the sequence of $n+1$ trials. So, to calculate $P (n+1 )$ from $ P (n) $, we simply have to add the probability that the triple in a row is not obtained until a final sequence NYYY. Note that the N preceding the triple YYY is strictly necessary to satisfy the condition that the triple has not been obtained in the first $n $ trials.
We can then start with $n=3$, for which we trivially have $P (3)=p^3$. For $n=4$, we have to add to $P (3) $ the probability to get a sequence NYYY, i.e. we have to add $p^3(1-p)$. Similarly, for $n=5$, we have to add to $P (4) $ the probability to get a final sequence NYYY (note that this is not affected by what happens in the first of the five trials), which means that we must add $p^3(1-p)$ as in the previous case. Lastly, for $n=6$, we have again to add to $P (5) $ the probability to get a final sequence NYYY (note that, as above, this is not affected by what happens in the first and in the second of the six trials). This leads to a further increase by $p^3(1-p)$. So, for this initial group of four values of in our probability progression, we directly obtain the recurrence
$$P (n+1)=P (n)+ p^3 (1-p)\,\,\,\, \text {(for n=3 to 6)}$$
Accordingly, for the specific case $p=0.7$, this formula gives the initial values
\begin{aligned} P(3) & =0.7^3=0.343 \\
P (4) & = 0.7^3+0.3 \cdot 0.7^3 = 0.4459 \\
P (5) & = 0.7^3+2\cdot 0.3 \cdot 0.7^3 = 0.5488 \\
P (6) & = 0.7^3+3\cdot 0.3 \cdot 0.7^3 = 0.6517 \\
\end{aligned}
already obtained in a previous answer via a different method.
Now let us consider the values of $n \geq 7$. Here something changes, because the fact that the sequence ends with NYYY does not imply that the triple YYY has not occurred before, i.e. in the first $n-4$ trials. So, at each step, the probability of getting a final NYYY must be multiplied to the probability of failure in the first $n-4$ trials, which is $1-P (n-4 ) $. Therefore, for $n=7$ we have to add to $P (6) $ the probability given by $p^3(1-p) [1-P(3)]\,\,$, for $n=8$ we have to add to $P (7) $ the probability given by $p^3(1-p) [1-P(4)] \,\, $, and so on. This leads to the recurrence
$$P (n+1)=P (n)+ p^3(1-p)[1-P(n-4)]\,\,\,\, \text{(for n}\, \geq \text {7)}$$
We can now try to get a closed form for this last recurrence. Let us set, by simplicity, $p^3=j$ and $(1-p) =k \,\, $. The recurrence becomes
$$P (n+1)=P (n)+ jk-jk \,P(n-4)\, \,$$
Note that the four values of $P (n)$ already determined can be written as $$P (3)=j $$
$$P (4)=j +jk$$
$$P (5)=j +2jk$$
$$P (6)=j +3jk$$
so that they contain only terms of degree $j $ or $jk $. Continuing our probability progression and focusing on the second group of four values of $n$ (from $n=7\,\, $ to $n=10 \,\, $), new terms of degree $j^2k $ and $j^2k^2$ appear applying the recurrence above (for the sake of readability, here I directly provide the resulting expressions avoiding the rather tedious step-by-step calcuations):
\begin{aligned} P (7) & = P(6)+ jk-jk P (3)=j+4jk-j^2k \\
P (8) & = P(7)+ jk-jk P (4)= j+ 5jk -2j^2k-j^2k^2 \\
P (9) & = P(8) + jk-jk P (5)= j+ 6jk -3j^2k-3j^2k^2 \\
P (10) & = P (9) + jk-jk P (6)= j+ 7jk -4j^2k-6j^2k^2 \\
\end{aligned}
Some pattern begins to emerge in these expansions. For the specific case $p=0.7$,
$$P (7)\approx 0.7193$$
$$P (8)\approx 0.7763$$
$$P (9) \approx 0.8228$$
$$P (10) \approx 0.8586$$
We can repeat the procedure, applying the same recurrence and continuing with the third group of four values of $n $ (i.e., $n=11$ to $14$). Again, new terms appear, in this case of degree $j^3k^2$ and $j^3 k^3$ (as above, I directly provide the resulting expressions):
\begin{align*} P(11) &=j+8jk-5j^2k-10j^2k^2+j^3k^2 \\ P(12) &=j+9jk-6j^2k-15j^2k^2+3j^3k^2+j^3k^3 \\P(13) & = j+10jk-7j^2k -21j^2k^2+6j^3k^2+4j^3k^3 \\
P (14) & = j+11jk-8j^2k-28j^2k^2+10j^3k^2+10j^3k^3\\
\end{align*}
The final pattern is therefore as follows. Our expansions of $P (n) $ include:
in the odd positions, $\lfloor (n+1)/4 \rfloor \,\, $ terms of the form $$(-1)^{i-1} \binom{n-3i}{i-1} j^i k^{i-1} \,\, $$ where $$i=1,2... \lfloor (n+1)/4 \rfloor \,\, $$ For example, in the expansion of $P (14)$ there are $ \lfloor (14+1)/4 \rfloor =3 \, \,\, $ terms of this form, which are $j $, $-8j^2k $, and $10 j^3k^2$;
in the even positions, $\lfloor n/4 \rfloor \,\, $ terms of the form $$(-1)^{i-1} \binom{n-3i}{i} j^i k^i \,\, $$ where $$i=1,2... \lfloor n/4 \rfloor$$
Accordingly, taking again for example the expansion of $P (14)$, there are $ \lfloor 14/4 \rfloor =3 \,$ terms of this form, which are $11jk$, $-28j^2k^2$, and $10 j^3k^3$.
Therefore, we obtain the following general expression for $P (n) $:
$$P (n)=\sum_{i}^{\lfloor (n+1)/4 \rfloor } (-1)^{i-1} \binom{n-3i}{i-1} j^{i}k^{i-1} +\sum_{i}^{\lfloor n/4 \rfloor } (-1)^{i-1} \binom{n-3i}{i} j^i k^i $$
Reminding that $j=p^3$ and $k=(1-p) $, we finally get
$$P (n)=\sum_{i}^{\lfloor (n+1)/4 \rfloor } (-1)^{i-1} \binom{n-3i}{i-1} p^{3i}(1-p)^{i-1} + \sum_{i}^{\lfloor n/4 \rfloor } (-1)^{i-1} \binom{n-3i}{i} p^{3i} (1-p)^i $$
Setting $p=0.7$, we obtain the probability asked in the OP for, which is
$$P (n)=\sum_{i}^{\lfloor (n+1)/4 \rfloor } (-1)^{i-1} \binom{n-3i}{i-1} 0.7^{3i} \cdot 0.3^{i-1} + \sum_{i}^{\lfloor n/4 \rfloor } (-1)^{i-1} \binom{n-3i}{i} 0.7^{3i}\cdot 0.3^i $$
Also note that the two sums have the same number of terms when $n \not\equiv -1 \,\, \text {mod}\,\, 4$, whereas the first sum has one term more than the second one when $n \equiv -1 \,\,\text {mod} \, 4$, because in this case $(n+1)/4$ is integer. This term is given by
$$\displaystyle (-1)^{\frac{(n+1)}{4}-1} \binom{n-\frac{3(n+1)}{4}}{ \frac{(n+1)}{4}-1}\, 0.7^{\frac{3(n+1)}{4}} \, 0.3^{\frac{(n+1)}{4}} $$
$$\displaystyle = (-1)^{\frac{(n-3)}{4}} \binom{\frac{(n-3)}{4}}{\frac{(n-3)}{4}}\, 0.7^{\frac{3(n+1)}{4}} \, 0.3^{\frac{(n+1)}{4}} $$
$$\displaystyle = (-1)^{\frac{(n-3)}{4}} \, 0.7^{\frac{3(n+1)}{4}} \, 0.3^{\frac{(n+1)}{4}} $$
So the probability asked in the OP can be alternatively expressed as
$$P (n)=\sum_{i}^{\lfloor n/4 \rfloor } (-1)^{i-1} \binom{n-3i}{i}\, 0.7^{3i}\, 0.3^{i} \left( \frac {i}{0.3(n-4 i+1)} +1 \right) + S$$
where
$$S = \begin{cases} 0 & \mbox{if } n \not\equiv -1 \, \text {mod}\, 4\\
(-1)^{\frac{(n-3)}{4}} \, 0.7^{\frac{3(n+1)}{4}} \, 0.3^{\frac{(n+1)}{4}} & \mbox{if } n \equiv -1 \, \text {mod}\, 4 \end{cases}$$
It could also be observed that $|S|$ is relatively small and rapidly decreases for increasing $n$: its value is $\approx 0.0106\, \,\, $ for $n=7$, becoming $\approx 0.00109\, \,\, $ for $n=11$ and $\approx 0.000112\,\,\, $ for $n=15$, so that its contribute to the overall probability is nearly negligible. The last expansion without the $S$ term can therefore considered a valid approximation even for the case $n \equiv -1 \, \text {mod}\, 4$.
The probability of three events in a row given by these formulas for $p=0.7$ and $n=3$ to $10$ have already been provided above. As expected, for increasing $n $, the value of $P (n)$ approaches $1$. The first value for which the probability achieves $95\%$ is $n=15$, since $P (15) \approx 0.9553\,\,\, $, as confirmed by WA here. The probability achieves $97.5\%$ and $99\%$ for $n=18$ and $n=22$, respectively, since $P(18) \approx 0.9772\,\,\, $ and $P(22) \approx 0.9909\,\,\, $, as confirmed by WA here and here.
