Answer: 14 zeroes.

First solution. Derive the maximum power of 2 as a common factor:
$$A = 2^7 \cdot (7^{14} + 1 + 2 \cdot 5^2 \cdot 7^{11} + 2^3 \cdot 5^4 \cdot 7^7 + 2^3 \cdot 5^6 \cdot 7^3).$$
Note that $50 = 7^2 + 1$ and represent 50 in this way in all terms:
$$2^{-7}A = 7^{14} + (7^2 + 1) \cdot 7^{11} + 2 \cdot (7^2 + 1)^2 \cdot 7^7 + (7^2 + 1)^3 \cdot 7^3 + 1.$$
Expand the right-hand side in powers of 7:
$$2^{-7}A = 7^{14} + 7^{13} + 7^{11} + 2 \cdot 7^{11} + 4 \cdot 7^9 + 2 \cdot 7^7 + 7^9 + 3 \cdot 7^7 + 3 \cdot 7^5 + 7^3 + 1.$$
Once again, use equality $7^2 + 1 = 50$, for which derive even degrees of 7 in all terms:
$$2^{-7}A = 7^{14} + 7 \cdot 7^{12} + 7 \cdot 7^{10} + 14 \cdot 7^{10} + 28 \cdot 7^8 + 14 \cdot 7^6 + 7 \cdot 7^8 + 21 \cdot 7^6 + 21 \cdot 7^4 + 7 \cdot 7^2 + 1$$
and expand the right-hand side in even powers of 7:
$$2^{-7}A = 7^{14} + 7 \cdot 7^{12} + 21 \cdot 7^{10} + 35 \cdot 7^8 + 35 \cdot 7^6 + 21 \cdot 7^4 + 7 \cdot 7^2 + 1.$$
Now it is clear that the right-hand side is the Newton binomial formula for $(7^2 + 1)^7$, hence the number $A = 2^7 \cdot 50^7 = 10^{14}$ ends with 14 zeroes.

Second solution For any $x_1$ and $x_2$ set $S_k = x_1^k + x_2^k$, $x_1 + x_2 = p$ and $x_1 x_2 = q$ (note that $S_0 = 2$, $S_1 = p$). Using the obvious recurrence formula $S_{k+2} = pS_{k+1} - qS_k$, we successively get $S_2 = p^2 - 2q$, $S_3 = p^3 - 3pq$, $S_4 = p^4 - 4p^2q + 2q^2$, $S_5 = p^5 - 5p^3q + 5pq^3$, $S_6 = p^6 - 6p^4 + 9p^2q^2 - 2q^3$ and, finally,
$$S_7 = p^7 - 7p^5q + 14p^3q^2 - 7pq^3 \quad (1)$$
Now, take $x_1 = 100$, $x_2 = -2$ then $p = 2 \cdot 7^2$, $q = -2 \cdot 10^2$ and (1) takes the form $10^{14} - 2^7 = 2^7 \cdot 7^{14} + 2^6 \cdot 7^{11} \cdot 10^2 + 2^6 \cdot 7^7 \cdot 10^4 + 2^4 \cdot 7^3 \cdot 10^6$ which is equivalent to $A = 10^{14}$.