Which stable homotopy groups are represented by parallelizable manifolds?

I think all elements are representable by honestly framed manifolds.

Let $M$ be a closed $d$-manifold with a stable framing, and consider the obstructions to destabilising a stable framing. Asumng $M$ is connected, which we can arrange by stably-framed surgery, there is a single obstruction, lying in $H^d(M ; \pi_d(SO/SO(d)))$.

If $d$ is even then $\pi_d(SO/SO(d)) = \mathbb{Z}$ and this obstruction may be identified with half the Euler characteristic of $M$. (As $M$ is stably framed, its top Stiefel--Whitney class vanishes and so its Euler characteristic is even.) We can change $M$ to $M \# S^p \times S^{2n-p}$ by doing a trivial surgery in a ball, and the stable framing extends over the trace of such a surgery. By taking $p$ to be 1 or 2 we can therefore change the Euler characteristic by $\mp 2$: thus we can change $M$ by stably framed cobordism until its Euler characteristic is 0, whence the stable framing destabilises to an actual framing.

If $d$ is odd then then $\pi_d(SO/SO(d)) = \mathbb{Z}/2$ and the obstruction is obscure to me (it is realised by the stable framing induced by $S^d \subset \mathbb{R}^{d+1}$, and is non-trivial even in Hopf invariant 1 dimensions where $S^d$ does admit a framing). I can't see an elementary argument for $d$ odd, but I think it is nontheless true by the following.

Let $d=2n+1$ with $d \geq 7$ (lower dimensions can be handled manually). Consider the manifold $$W_g^{2n} = \#g S^n \times S^n.$$ This has a stable framing by viewing it as the boundary of a handlebody in $\mathbb{R}^{2n+1}$. By doing some trivial stably-framed surgeries as above (with $p=2,3$ say, to keep it simply-connected), we can change it by a cobordism to a manifold $X$ having an honest framing $\xi$. I wish to apply [Corollary 1.8 of Galatius, Randal-Williams, ``Homological stability for moduli spaces of high dimensional manifolds. II"], to $(X, \xi)$. There is a map $$B\mathrm{Diff}^{fr}(X, \xi) \to \Omega^{\infty+2n} \mathbf{S}$$ given by a parameterised Pontrjagin--Thom construction. Now there is a step that I would have to think about carefully, but I think that the choices made can be arranged so that $(X,\xi)$ has genus $g$ in the sense of that paper, and so taking $g$ large enough the map above is an isomorphism on first homology. But this has the following consequence: any element $x \in \pi_{2n+1}(\mathbf{S})$ is represented by the total space of a fibre bundle $$X \to E^{2n+1} \overset{\pi}\to S^1$$ with a framing of the vertical tangent bundle (and the Lie framing of $S^1$).

(Again, I'm sure there must be a more elementary way of seeing this.)

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:

The cases $n=1,3,7$ are fine, since the stable stems are in these degrees 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).

$k \cdot[\mathrm{point}]\in \pi_0^s$ is represented by an honestly framed 0-manifold if and only if $k \geq 0$.