Solution:

Every term is a power of $3$ and the ratio is a quotient of powers of $3,$ so $x_n = 3^{a + bn}$ for integers $a$ and $b,$ with $b \ge 1$ since the sequence increases. The first condition gives $$\sum_{n=0}^{7} (a + bn) = 8a + 28b = 308, \qquad \text{i.e.} \qquad 2a + 7b = 77.$$

For the second condition, $x_7 = 3^{a+7b}$ is the largest term, and $$x_7 \lt \sum_{n=0}^{7} x_n \lt x_7\left(1 + \tfrac{1}{3} + \tfrac{1}{9} + \cdots\right) = \tfrac{3}{2}\,x_7 \lt 3x_7,$$ so $\log_3\left(\sum x_n\right)$ lies strictly between $a + 7b$ and $a + 7b + 1.$ The given bounds then force $a + 7b = 56.$

Subtracting from $2a + 7b = 77$ yields $a = 21,$ then $b = 5.$ Therefore $\log_3(x_{14}) = a + 14b = 21 + 70 = 91.$