$\boxed{2042 \cdot 19^{14}}$

First we show the following lemma.

Lemma. Let $a, b, c$ be integers satisfying $a \neq 0 \pmod{17}$, $b \ge 1$, $c \ge 4$. Then $a^{b^c} \equiv 1 \pmod{17}$ holds if and only if $a \equiv 1 \pmod{17}$ or $b$ is even.

Proof. If $a \equiv 1 \pmod{17}$, obviously $a^{b^c} \equiv 1 \pmod{17}$ holds. If $b$ is even, $c \ge 4$ shows $b^c$ is divisible by $16$. By Fermat's little theorem $a^{16} \equiv 1 \pmod{17}$ holds thus $a^{b^c} \equiv 1 \pmod{17}$. We show the converse. If $a^{b^c} \equiv 1 \pmod{17}$ holds, we can take minimum positive integer $d$ satisfying $a^d \equiv 1 \pmod{17}$. If $a \neq 1 \pmod{17}$ then $d \neq 1$. Assume $b^c$ is not divisible by $d$, then $b^c$ can be written in the form $b^c = sd + t$ with a non-negative integer $s$ and an integer $t$ satisfying $1 \le t < d$ hence
$$a^t \equiv (a^d)^s \cdot a^t = a^{b^c} \equiv 1 \pmod{17},$$
which contradicts to the minimality of $d$. Therefore $b^c$ is divisible by $d$. Similarly $16$ is divisible by $d$ and $d \neq 1$ shows $d$ is even. Hence $b^c$ is even and thus $b$ is even. ■

If $a_1 \equiv 0 \pmod{17}$ or $a_2 \equiv 0 \pmod{17}$ then the condition is not satisfied. From here we assume $a_1 \neq 0 \pmod{17}$ and $a_2 \neq 0 \pmod{17}$. Let
$$c_1 = a_3^{a_4^{a_5^{a_6^{a_7}}}}, \quad c_2 = a_4^{a_5^{a_6^{a_7}}},$$
then $c_1 \ge 2^2 = 4$, $c_2 \ge 2^2 = 4$ holds. We have two cases; when $a_2$ is odd and when it is even.

* When $a_2$ is odd.

Lemma shows $a_1^{a_2^{c_1}} \equiv 1 \pmod{17}$ if and only if $a_1 \equiv 1 \pmod{17}$. Since $a_2 \neq 18$, we have $a_2 \neq 1 \pmod{17}$ and then lemma shows $a_2^{a_3^{c_2}} \equiv 1 \pmod{17}$ if and only if $a_3$ is even. Therefore the number of tuples satisfying the condition is $1 \cdot 8 \cdot 10 \cdot 19^{14} = 80 \cdot 19^{14}$.

* When $a_2$ is even.

We have $a_1 \neq 0 \pmod{17}$ then lemma shows $a_1^{a_2^{c_1}} \equiv 1 \pmod{17}$. By lemma, $a_2^{a_3^{c_2}} \equiv 1 \pmod{17}$ holds if and only if $a_2 \equiv 1 \pmod{17}$ or $a_3$ is even. Therefore the number of tuples satisfying the condition is $18 \cdot 1 \cdot 19^{15} + 18 \cdot 9 \cdot 10 \cdot 19^{14} = 1962 \cdot 19^{14}$.

Hence the total number of tuples satisfying the condition is $80 \cdot 19^{14} + 1962 \cdot 19^{14} = 2042 \cdot 19^{14}$.