There is an old reference which gives low-dimensional examples. The paper

A. Dold and H. Whitney. Classification of oriented sphere bundles over a $4$-complex. Ann. Math.(2) 69 (1959), pp. 667--677

contains a classification of oriented vector bundles over CW-complexes of dimension $4$. Take a rank $n$ real vector bundle $\mathcal{E}$, $n\geq 5$, with trivial Stiefel-Whitney classes and trivial Pontryagin class over a $4$-dimensional CW-complex $X$. Then there is a cohomology class in $H^4(X,\mathbb{Z})$ which measures the difference between $\mathcal{E}$ and the trivial bundle, the class is $0\mod 2$ and annihilated by multiplication by $2$. Conversely, any such class can be realized by a bundle.

Therefore, a non-trivial bundle exists on a $4$-complex if there are classes which are $0\mod 2$ and annihilated by $2$. The obvious example of such a complex is the Moore space $M(\mathbb{Z}/4,4)$ obtained by glueing a $4$-cell to $S^3$ with a degree $4$ map. Note also that rank $n\geq 5$ on a $4$-complex is in the stable range, so the bundle obtained from the Dold-Whitney paper is stably non-trivial.

Edit: Since the question was tagged differential-geometry, I'll also give a manifold example. Take a free action of $\mathbb{Z}/4\mathbb{Z}$ on $S^5$ (which can be obtained from the embedding $S^5\subset \mathbb{C}^3$ and coordinatewise multiplication with suitable roots of unity). The quotient $X=S^5/(\mathbb{Z}/4\mathbb{Z})$, a lens space, is a $5$-dimensional smooth manifold with $H^4(X,\mathbb{Z})\cong\mathbb{Z}/4\mathbb{Z}$. This means that there is a non-trivial $SO(6)$-bundle with trivial characteristic classes on the $4$-skeleton, as follows from the Dold-Whitney paper. Since $\pi_5(BSO(6))=0$, obstruction theory implies that any $SO(6)$-bundle on the $4$-skeleton extends uniquely to a vector bundle on $X$. Again, rank 6 on dimension 5 is stable range, so the bundle constructed this way is stably non-trivial.