It is known that the probability of a gaussian random matrix being full rank is 1 (e.g. here: Probability that a random matrix will have full column rank?).
Therefore the probability of a gaussian random vector (i.i.d gaussian entries) being a linear combination of other fixed vectors is should also be zero. I have a more specific subset, I want to show that the probability of the vector being in the set of all combinations of other fixed vectors so that the weights add up to 1 is zero. So I just tried to prove it for $d=2$ (dimensionality) and $n=2$ fixed vectors. I must have made a fundamental error, because it's not zero.
\begin{align} &\mathcal{P}(X = \alpha \cdot w_1 + (1-\alpha) \cdot w_2, \alpha \in \mathbb{R}) \\ &=\int_{\alpha \in \mathbb{R}}f_{X_1}(\alpha \cdot [w_1]_1 + (1-\alpha) \cdot [w_2]_1) \cdot f_{X_2}(\alpha \cdot [w_1]_2 + (1-\alpha) \cdot [w_2]_2) d \alpha \\ &=\int_{\alpha \in \mathbb{R}}f_{\mathcal{N}(0,1)}(\alpha \cdot [w_1]_1 + (1-\alpha) \cdot [w_2]_1) \cdot f_{\mathcal{N}(0,1)}(\alpha \cdot [w_1]_2 + (1-\alpha) \cdot [w_2]_2) d \alpha \\ &=\frac{1}{2\cdot \pi}\int_{\alpha \in \mathbb{R}}e^{-\frac{1}{2}((\alpha \cdot [w_1]_2 + (1-\alpha) \cdot [w_2]_2)^2 + (\alpha \cdot [w_1]_2 + (1-\alpha) \cdot [w_2]_2)^2)} d \alpha \\ &=\frac{1}{2\cdot \pi}\int_{\alpha \in \mathbb{R}}e^{-\frac{1}{2}([w_1]_1 - [w_2]_1)^2([w_1]_2 - [w_2]_2)^2\alpha^2} d \alpha \\ &> 0 \text{ since it's a gaussian integral} \end{align}
where am I wrong?
The set $\{ \alpha w_1 + (1-\alpha)w_2 : \alpha \in \mathbb R\}$ is exactly the line passing through the points $w_1$ and $w_2$ since $$\alpha w_1 + (1-\alpha)w_2 = \alpha(w_1-w_2) + w_2.$$
This line $L$ has no area as subset of $\mathbb R^2$ so by definition the integral of any function on $L$ is equal to $0$.
Concerning your calculuations you should be careful. It is not because you have a parametrization $\gamma : \mathbb R \to L : \alpha \mapsto \alpha w_1 + (1-\alpha w_2)$ with $\gamma (\mathbb R) = L$ that you can write
$$ \int_{\gamma(\mathbb R)} f(x) dx = \int_{\alpha \in \mathbb R} f \circ \gamma(\alpha ) d\alpha$$
This is not at all how we do changes of variables in multivariable calculus !