Assume that $n(t)$ is a White Gaussian Noise (WGN) process with $E[n(t)]=0$, $E[n(t)^2]=\sigma^2$ and $f(t)$ a deterministic function defined in $[0,T]$. How can I compute from first principles the variance of $g(T)$ defined as
$$g(T)=\int_0^Tf(t)n(t)dt.$$
Any references to elementary textbooks on stochastic processes are also welcome.
White noise can only be defined in the sense of distributions or as a measure. A good definition can be found in Adler and Taylor (2007, Sec. 1.4.3), see also this SE answer. If it would make sense the relation of this measure definition and your $n$ would be
$$ n(t) = \lim_{\epsilon\to 0}\frac{W([t,t+\epsilon))}{\epsilon} $$
To calculate second moments you want to use stochastic integration Adler and Taylor (2007, sec. 5.2) for deterministic functions $f,g$ $$ \mathbb{E}[W(f)W(g)] \overset{\text{def.}}=\mathbb{E}\Bigl[\Bigl(\int f(x)W(dx)\Bigr)\Bigl(\int g(x)W(dx)\Bigr) \Bigr] =\int f(x)g(x) dx, \tag{1} $$ which can be viewed as a special case of the Itô Isometry.
In your case you are interested in $$ g(T)=W(f\mathbf{1}_{[0,T)})=\int_0^T f(t) W(dt) = \langle f\mathbf{1}_{[0,T)}, W\rangle $$ for which you get the variance $$ E[g(T)^2] = \int_0^T f(t)^2 dt $$
The trick to prove (1), is to show that the mapping $$ W:\begin{cases} L^2(\mathbb{R}^n, \mathcal{B}, \nu) &\to L^2(\Omega, \mathcal{A}, \mathbb{P})\\ f &\mapsto W(f) := \int f(t) W(dt) \end{cases} $$ preserves the scalar product. We first consider simple functions $f=\sum_{i=1}^n a_i \mathbf{1}_{A_i}$ for disjoint $A_i$, then $$ W(f)=\int f(t) W(dt) \overset{\text{def.}}= \sum_{i=1}^n a_i W(A_i) $$
Comment: in particular the expectation is zero and variance given by $\sum_{i=1}^n a_i \nu(A_i)$ considering the definition of $W$.
To calculate the scalar product between $f$ and $g=\sum_{i=1}^n b_i \mathbf{1}_{B_i}$ we assume without loss of generality $A_i=B_i$ (consider all of their interesections). Then $$\begin{aligned} \langle W(f), W(g) \rangle_{L^2(\mathbb{P})} &= \mathbb{E}\Bigl[\sum_{i=1}^n a_i W(A_i) \sum_{j=1}^n a_j W(A_j)\Bigr]\\ &= \sum_{i=1}^n a_i b_i \mathbb{E}[W(A_i)^2]\\ &= \int f(t) g(t) \nu(dt)\\ &= \langle f, g\rangle_{L^2(\nu)} \end{aligned}$$ Since the simple function are dense in $L^2$ and the scalar product is continuous we can deduce that $W$ is an isometry (where $W(f)$ for general $f$ is defined as the limit of $W(f_n)$ for simple functions $f_n$ approximating $f$). (1) then directly follows from the respective definitions of the scalar product (i.e. the second and penultimate term).