Vector bundle of dimension $\leqslant n$ on $n$-connected space is trivial

No, this is not the case. As Najib points out in the comments, this is the same as asking "if $X$ is $n$-connected, is every map $X \to BO(n)$ null-homotopic?"

Instead, let's show this isn't true for oriented bundles; if every bundle is (unoriented) trivial, then there are 2 oriented bundles up to isomorphism.

Take $X = S^4$ and $n = 3$. $SO(3)$ has a double cover isomorphic to the group of unit quaternions $S^3$, which fits into a fibration $\Bbb Z/2\Bbb Z\to BSO(3) \to BS^3$, and $BS^3 = \Bbb{HP}^\infty$ (this coming from a more general fibration $G/H \to BH \to BG$). Passing to the long exact sequence of homotopy groups we see $\pi_k(BSO(3)) = \pi_k(\Bbb{HP}^\infty)$. By cellular approximation, $\pi_4(\Bbb{HP}^\infty) = \pi_4(S^4) \cong \Bbb Z$, which gives far more oriented 3-bundles over $S^4$ than possible if they were all (unoriented) trivial.

Your proposed argument fails at the third step. It is just not true that if $X$ is $n$-connected then every $S^{n-1}$-bundle over it is trivial.

The desired statement is not even true rationally: the rational homotopy groups of $BO(n)$ are not hard to calculate and in particular for $n \ge 3$,

$$\pi_{4n-4}(BO(2n)) \otimes \mathbb{Q} \cong \mathbb{Q}$$

so it follows that there are countably many nontrivial $2n$-dimensional real vector bundles over $S^{4n-4}$, which moreover can be distinguished by their Pontryagin classes $p_{n-1}$, and this gives an infinite family of counterexamples to the desired statement.