This question is motivated by an observation of Milnor.
Theorem: Let $M^m$, $N^n$ be parallelizable smooth manifolds, and $i:M\to N$ an embedding. If $n>2m$, the normal bundle is trivial.
The proof uses the fact that the Stiefel manifolds are highly connected, so that we rearrange our global sections of N in a neighborhood of $i(M)$ to match the sections of $TM$. This argument is given on page 48 of Milnor, A procedure for killing homotopy groups of differentiable manifolds.
I am interested in a counterexample for the case of low codimension.
One of the easiest places to find parallelizable manifolds is 3-manifolds: oriented ones are parallelizable. So as Qiaochu says in his comment, we should consider cases which have stably trivial normal bundle. So why not consider stably trivial bundles over 3-manifolds? The only place you can find this in the dimension you consider is $SO(2)$-bundles; the Euler class is not stable. Indeed $SO(2)$-bundles over 3-manifolds are stably trivial iff $w_2 = 0$; that is, if the Euler class is a multiple of 2. (This is because of the Dold-Whitney theorem that $SO(3)$-bundles over 4-complexes are determined by $w_2$ and $p_1$.)
I claim that the total space of any such bundle is parallelizable. It actually suffices to produce a frame on the zero section, for then we can extend it to a small neighborhood of the zero section. This small neighborhood is diffeomorphic to the whole normal bundle. (Better: the tangent bundle to a vector bundle is the pullback under the multiplication-by-0 map of its restriction to the 0 section.)
But $TE \cong TM \oplus E$ on the zero section, and $E$ is stably trivial, so $TE$ is trivial. This proves that the total space of this vector bundle is parallelizable. Because the normal bundle itself was not trivial, this provides an example. We have thus constructed many examples; one for every oriented 3-manifold $M$ with a nontrivial element in $2H^2(M;\Bbb Z)$.