So to avoid a merely opinion-based answer, let us examine firstly the entropy of some variable $x \sim \mathcal{N}(\mu, \sigma^2 )$.
$$H(x)=-\int p(x) \log p(x) = -\mathbb{E} \big[\log \mathcal{N}(\mu, \sigma^2) \big] \tag{1, 2}$$ $$= - \mathbb{E}\Bigg[\log \Big[(2\pi\sigma^2)^{-\frac{1}{2}} e^{-\frac{1}{2\sigma^2}(x-\mu)^2} \Big] \Bigg] \tag{3}$$ $$=\frac{1}{2}\log(2\pi\sigma^2)+\frac{1}{2\sigma^2} \mathbb{E}\big[(x-\mu)^2 \big]\tag{4}$$ $$=\frac{1}{2}\log(2\pi\sigma^2)+\frac{1}{2} \tag{5}$$
(To understand the transition from $(4)$ to $(5)$ simply remember the definition of variance: $\mathbb{E}\big[(x-\mu)^2 \big]=\sigma^2$.)
As you see, the entropy of a normal distribution is a function of its variance. That is intuitive, because high variance implies higher entropy and vice versa. But what is this telling us?
Firstly, consider how we are defining information:
$$H(X)=\mathbb{E}_{x\sim p}[I(x)] = -\mathbb{E}_{x\sim p}[\log P(x)] = -\int_{-\infty}^\infty p(x)\log p(x)$$
This is a definition that formalizes the notion that unlikely events convey more information than likely events. This formal definition does not always align with more common-sense notions of information (for example, the amount of conclusions we can draw from a certain data). In fact, they may even be contradictory.
For example, imagine two variables $X \sim \mathcal{N}(\mu_1, \sigma_1^2)$ and $Y\sim \mathcal{N}(\mu_2, \sigma_2^2)$ with $H(X)$ very high and $H(Y)$ very low. $X$ conveys more information, in the formal sense of the definition. However, this is not without problems. The most elementary example is that $\mu_1$ will be far less representative of $X$ than $\mu_2$ of $Y$. At the same time, if you model data with a Gaussian distribution and your model has very high entropy, this may be an indicator of overfitting. In fact, Gaussian models are very prone to overfitting in general.
As you can see, high entropy is not necessarily equal to statistical usefulness. So $1)$ why is entropy valued and $2)$ why is the normal distribution so important?
Maximum entropy is valuable mainly because when we do not know the real function $f$ that defines some data $X$ we want to find a model $\hat{f}$ that has the least possible assumptions. As explained in one of your past questions, high entropy does guarantee this (it is simply entailed in the definition). (Again, see Occam's razor).
The normal distribution is a maximum entropy distribution. That has value with regards to point $1)$, but is not good enough to answer our question. If we couldn't draw important conclusions from a Gaussian distribution, the fact that it is safe to assume such distribution would not be very useful. The normal distribution has properties not related to its entropy that make it extremely valuable, but I will not enumerate them because they are basic and well-known. (Simply consider the value in the central limit theorem!) Because of these properties, a normal distribution is very nice to work with. One can generally draw stronger conclusions with less effort from a normal distribution than from a non-normal distribution (hence the value in normalizing non-normal variables).
So my final answer is no, maximum entropy is not enough to explain the prevalence and popularity of the normal distribution in statistics. Arguably, the central limit theorem plays a greater role at that.
