Repeating the first part of Oscar's answer and elaborating on comments by Chris and Panagiotis, here is a down-to-earth argument in all cases:

These cases $n=1,3,7$ are fine, since in these cases the stable stems are generated by $S^1$, $S^3$, $S^7$ with the unstable framing induced by the multiplication in the unit complex numbers, quaternions, or octonions.

In the other cases, we use that the obstruction to destabilising a given stable framing $F$ of an oriented closed manifold $M^n$ lies in $H^n(M,\pi_n(SO/SO(d))$, which is isomorphic (in a preferred way) to $\mathbb{Z}$ if $n$ is even and to $\mathbb{Z}/2$ if $n$ is odd. It is not too hard to see that, with respect to this isomorphism, the obstruction is given by the *semi-characterstic*: half the Euler characteristic for $n=2d$ and $\sum_{i=0}^d\mathrm{dim}(H_i(M,\mathbb{Z}/2))\text{ mod }(2)$ for $n=2d+1$ and $n\neq1,3,7$. In particular, the obstruction to destabilising is independent of $F$ which is somewhat surprising.

Originally this was proved by to Bredon and Kosinksi [1] who used a more geometric description of this obstruction: it is the degree (mod $2$ if $n$ is odd) of the Gauss map $M\rightarrow{S^n}$ induced by the stable framing $TM\oplus \varepsilon\cong \varepsilon^{n+1}$ (take the image of the canonical vector field in the trivial line bundle and normalize).

Now observe that, as Oscar explained, by doing a couple of trivial surgeries in a ball corresponding to taking connected sums with $S^1\times S^{n-1}$ or $S^2\times S^{n-2}$ and extending the stable framing, any stably framed bordism class in even dimensions contains a representative with trivial Euler-characteristic. The same works with the semi-characteristic in odd dimensions (here at most one surgery is necessary), so by the discussion above every stably framed bordism class has a representative whose stable framing can be destabilised.

[1] G.E. Bredon and A. Kosinski, *Vector fields on $\pi$-manifolds*. Annals of Math. 84, 85– 90 (1960).