As explained at https://stats.stackexchange.com/a/314739/919, Fisher's Method combines p-values $p_1, p_2, \ldots, p_n$ under the assumption they arise independently under null hypotheses with continuous test statistics. This means each is independently distributed uniformly between $0$ and $1.$ A simple calculation establishes that $-2\log(p_i)$ has a $\chi^2(2)$ distribution, whence
$$P = \sum_{i=1}^n -2\log(p_i)$$
has a $\chi^2(2n)$ distribution. For large $n$ (as guaranteed by the Central Limit Theorem) this distribution is approximately Normal. It has a mean of $2n$ and variance of $4n,$ as we may readily calculate.
Suppose, now, that $P$ is "much" different than this mean. "Much" means, as is usual, in comparison to the standard deviation. In other words, suppose that $P$ differs from $2n$ by more than a few multiples of $\sqrt{4n}=2\sqrt{n}.$ From basic information about Normal distributions this implies that $P$ is either unusually small or unusually large. Consequently, as $P$ ranges from $2n-2K\sqrt{n}$ to $2n+2K\sqrt{n}$ for $K \approx 3,$ Fisher's method assigns a cumulative probability (that is, combined p-value) ranging from nearly $0$ to nearly $1.$
In other words, all of the "interesting" probability for $P$ occurs within the interval $(2n-2K\sqrt{n}, 2n+2K\sqrt{n})$ for small $K$. As $n$ grows, this interval narrows relative to its center (at $2n$).
One conclusion we could draw from this result is that when $\sqrt{n}$ is large enough to dominate $2K$--that is, when $n$ is much larger than $(2\times3)^2\approx 40$ or so, then Fisher's Method may be reaching the limits of its usefulness.
In the circumstances of the question, $n=10^7.$ The interesting interval for the average log p-value, $-P/(2n),$ therefore is roughly
$$-(2n-2K\sqrt{n}, 2n+2K\sqrt{n})/(2n) \approx (-0.999051, -1.00095)$$
when $K=3.$
The corresponding geometric mean p-values are
$$e^{-0.999051} = 0.368229\text { and } e^{-1.00095} = 0.367531.$$
The lower value of $0.367$ used in the question is outside this interval, giving essentially zero (lower) tail probability, while the upper value of $0.368$ lies within this interval, giving a probability that is still appreciably less than $1.$ This is an extreme example of our previous conclusion, which could be restated like this:
When the average natural logarithm of the p-values differs much from $-1,$ Fisher's Method will produce a combined p-value extremely near $0$ or near $1$. "Much" is proportional to $1/\sqrt{2n}.$