$H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group

I try to give an argument without spectral sequences, not sure if this can be considered non-computational though. At least, there is a non-computational syllabus: torsion classes in $H^4(BG,\mathbb{Z})$ would be characteristic classes of torsion bundles over $S^3$ but the latter have to be trivial.

Now for a slightly more detailed argument: first, the coefficient formula tells us that torsion in $H^4(BG,\mathbb{Z})$ comes from torsion in $H_3(BG,\mathbb{Z})$ because torsion in $H_4(BG,\mathbb{Z})$ would not survive the dualization. Since $G$ is connected, the Hurewicz theorem provides a surjection $\pi_3(BG)\to H_3(BG,\mathbb{Z})$, so any torsion class in $H_3(BG,\mathbb{Z})$ comes from a map $S^3\to BG$ classifying a $G$-bundle over $S^3$. The corresponding clutching map $S^2\to G$ is homotopic to the constant map because $\pi_2(G)$ is trivial. So we find that any $G$-bundle on $S^3$ is trivial, hence $H_3$ and therefore $H^4$ of $BG$ do not have non-trivial torsion.

Here is another spectral sequence argument. It is mainly of interest because it is different, but I think that it is slightly easier and will detail how. The spectral sequence is used by Deligne in the paper cited in the comments by Guest, but only for the simply connected case. The advantage is that André's argument required knowing that $H^4(K(\pi,2);\mathbb Z)$ is torsion-free, while this argument does not. Also, it does not seem to require knowing that $\pi_2$ of a Lie group vanishes or that $\pi_3$ is torsion-free. Indeed, it appears to prove those facts, but they are probably closely related to the required input. What it does require knowing is that $K/T$, the quotient of a compact group by its maximal torus, has cohomology that is torsion-free and concentrated in even degrees. This is usually shown by the Schubert (or Bruhat) decomposition into even dimensional cells. The spectral sequence is the Serre spectral sequence for the fiber sequence $K/T\to BT\to BK$, namely $H^*(BK; H^*(K/T))\Rightarrow H^*(BT)$. Since $BK$ is simply connected, the cohomology is untwisted; and since the coefficients are torsion-free (another advantage over the other spectral sequence), the $E_2$ becomes a tensor product: $H^*(BK)\otimes H^*(K/T)\Rightarrow H^*(BT)$ and takes this form: $$\begin{matrix} \vdots\\ * & \vdots \\ 0 & 0 \\ * & 0 & * \\ 0 & 0 & 0 & 0 \\ \mathbb Z & 0 & * & * & H^4(BK) & \cdots \\ \end{matrix}$$

There is no room for a differential to come from or hit $H^4(BK)$, so it injects into $H^4(BT)$, which is torsion-free. Injecting into a torsion-free group is slightly nicer than in the other spectral sequence, where it is an extension of two torsion-free groups. In the simply connected case, $H^4(BK)=H^4(BT)^W$, but that is not true in general (cf the spin characteristic class $\frac{p_1}2$). and that's also true in the non-simply connected case (cf comments below).