As you correctly say, if $\{X(t)\colon t \in \mathbb T\}$ is a Gaussian process, then for every possible choice of $n$ random variables $X(t_1), X(t_2), \ldots, X(t_n),$ where $t_k \in \mathbb T$ for $ k = 1,2,\ldots, n$, the $n$ variables are jointly Gaussian. Indeed, that is the definition of a Gaussian random process: a collection of random variables such that every finite subset of random variables is jointly Gaussian, that is, has a multivariate normal density. Note that the individual random variables are also themselves Gaussian random variables. Note also that the definition does not require that the index set $\mathbb T$ be a countable set such as $\mathbb Z$, the set to all integers (which would make the process a discrete-time process); having $\mathbb T = \mathbb R$, the set of real numbers, is fine (this makes the process a continuous-time process): but in either case,the requirement is still the same that every finite subset of random variables that you can choose from the process has a jointly Gaussian distribution. There is no such thing as an infinite-dimensional joint distribution function (joint Gaussian or otherwise), and so the definition is silent about what can be said about the joint distribution of infinite subsets of the random variables.
You are asking whether it is true that if we have
a collection of Gaussian random variables, specifically the set $\{X(t)\colon t \in \mathbb T\}$, such that each member $X(t)$ of the set is a Gaussian random variable,
then
is it always the case that the set constitutes a Gaussian process?
The answer is No if you accept the definition above because Gaussian random variables don't have to be jointly Gaussian, and is so there is no guarantee that the chosen collection of Gaussian random variables is also a collection of jointly Gaussian random variables, as required in the definition of Gaussian random process. An incredibly diverse collection of joint densities of two Gaussian random variables that are not jointly Gaussian can be found in this answer on stats.SE.
Turning to your "micro phenomenon" comment, it seems that you are (or the author of your textbook is) thinking that the assumption of individual Gaussianity is justified from a central limit theorem type view of the distribution of each individual random variable. This is not quite correct. The Gaussian random variable model usually comes from an assumption that one is dealing with thermal noise which is caused by the random motion of electrons inside a conductor. The (small) charge on each electron induces a (small) voltage across the terminals of the conductor depending on the position of the electron with respect to the terminals, and all the zillions of such voltages add up (or cancel out) to produce a net voltage which can be modeled a Gaussian random variable. This net voltage varies with time as the electrons jostle about, creating a random process that is generally modeled as a Gaussian process, that is, we make a leap of faith that the random variables are not just individually Gaussian (which we can justify as above in a hand-waving sort of way) but also jointly Gaussian. One justification for this leap of faith is that if we imagine a linear system with white Gaussian noise as input, then the output process is a Gaussian process. More important, the linear system with white Gaussian noise input model has considerable support from experimental evidence it matches up well with physical reality.
@AlexTP Uncorrelated Gaussian random variables are not necessarily jointly Gaussian random variables. The classic example is $X \sim N(0,1), Y = ZX$ where $Z$ takes on values $\pm 1$ with equal probability $\frac 12$. Then, $Y \sim N(0,1)$ also. Furthermore, $X$ and $Y$ are uncorrelated random variables, but $X$ and $Y$ are not jointly Gaussian random variables. In fact, all the probability mass lies along two straight lines intersecting at right angles instead of being distributed with circular symmetry in the plane as happens with jointly Gaussian random variables with equal variances.
