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Added$\strut^2$: The Euler characteristic is an obstruction to quaternionic structure when $n\geq 2$. In particular this proves that $\mathbb{C}P^{2n}$ does not have quaternionic structure for $n\geq 2$.

proof: Above we argued that the images of the maps $\pi_{4n}(B\mathrm{Sp}(n)) \to \pi_{4n}(B\mathrm{SO}(4n))$ and $\pi_{4n}(B\mathrm{Sp}(1)\times B\mathrm{Sp}(n)) \to \pi_{4n}(B\mathrm{SO}(4n))$ is the same for $n\geq 2$. The first map (and thus both) is detected not to be surjective by the Euler charateristic given by

$$\pi_{4n}(B\mathrm{SO}(4n)) \xrightarrow{h} H_{4n}(B\mathrm{SO}(4n)) \xrightarrow{e} \mathbb{Z}$$

Here $h$ is the Hurewitz map, and $e$ is given by capping with the Euler class of the canonical bundle. Detection here means that $e$ is zero on the images. So by considering homotopy groups we see that if a lift exist then the Euler class must be zero.

In fact up to 2-torsion the Euler class is the only obstruction. Indeed, the inclusions above are mod 2-torsion and this copy of $\mathbb{Z}$ and equivalence.

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I think this is true in general. Indeed, it is true for $n=1$ where the above is not a contradiction because there $\pi_3(\mathrm{Sp}(1))\cong \mathbb{Z}$. Andrei pointed out in a comment that this is also true for $n=1,2$.

Added$\strut^2$: The Euler characteristic is an obstruction to quaternionic structure when $n\geq 2$. In particular this proves that $\mathbb{C}P^{2n}$ does not have quaternionic structure for $n\geq 2$.

proof: Above we argued that the images of the maps $\pi_{4n}(B\mathrm{Sp}(n)) \to \pi_{4n}(B\mathrm{SO}(4n))$ and $\pi_{4n}(B\mathrm{Sp}(1)\times B\mathrm{Sp}(n)) \to \pi_{4n}(B\mathrm{SO}(4n))$ is the same for $n\geq 2$. The first map (and thus both) is detected not to be surjective by the Euler charateristic given by

$$\pi_{4n}(B\mathrm{SO}(4n)) \xrightarrow{h} H_{4n}(B\mathrm{SO}(4n)) \xrightarrow{e} \mathbb{Z}$$

Here $h$ is the Hurewitz map, and $e$ is given by capping with the Euler class of the canonical bundle. Detection here means that $e$ is zero on the images. So by considering homotopy groups we see that if a lift exist then the Euler class must be zero.

In fact up to 2-torsion the Euler class is the only obstruction. Indeed, the inclusions above are mod 2-torsion and this copy of $\mathbb{Z}$ and equivalence.

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It seems to me that if I understood the comments to my comment correctly that the map

$$\mathrm{Sp}(1) \times \mathrm{Sp}(n) \to \mathrm{SO}(4n)$$

induced by right unit quarternionic multiplaction on $\mathbb{H}^n$ of the left factor and right matrix multiplication on $\mathbb{H}^n$ of the left factor has kernel $\{ \pm 1\}$. Since the source is simply connected it must lift to the spin group. So we have a map

$$\mathrm{Sp}(1) \times \mathrm{Sp}(n) \to \mathrm{Spin}(4n)$$.

Covering the map

$$\mathrm{Sp}(1)\mathrm{Sp}(n) \to \mathrm{SO}(4n)$$

Since the covering fiber is $\mathbb{Z}/2\mathbb{Z}$ and we can check that after taking the functor $B$ both fibers are $K(\mathbb{Z}/2\mathbb{Z},1)$-spaces we see that

$$\begin{matrix} B(\mathrm{Sp}(1)\times \mathrm{Sp}(n)) & \longrightarrow & B(\mathrm{Spin}(4n)) \\ \downarrow && \downarrow \\ B(\mathrm{Sp}(1)\mathrm{Sp}(n)) & \longrightarrow & B(\mathrm{SO}(4n)) \end{matrix}$$

is homotopy cartesian.

So if $M$ is spinable and has an almost Quarternionic structure it means that its classifying map lifts to $B(\mathrm{Sp}(1) \times \mathrm{Sp}(n))$

Edit: The conclusion (which is now removed) was wrong, but at least it seems to simplify the picture when $M$ is spin.

Added: For spheres $S^{4n}$ we may use the above on the $4n$th homotopy group and deloop. This implies that if we had a quartenionic structure on $S^{4n}$ we would have that the image of the map

$$\pi_{4n-1}(\mathrm{Sp}(1)\times \mathrm{Sp}(n) ) \to \pi_{4n-1} (\mathrm{SO}(4n))$$

contains the image of the map $\mathbb{Z} \cong \pi_{4n-1}(\Omega S^{4n}) \to \pi_{4n-1}(\mathrm{SO}(4n)) \cong \mathbb{Z}\times \mathbb{Z}$ (*) induced by the delooping of the classifying map for the tangent bundle of $S^{4n}$.

We know that not having a hyper kahler an almost hypercomplex structure implies that the image of $\pi_{4n-1}(\mathrm{Sp}(n)) \to \pi_{4n-1} (\mathrm{SO}(4n))$ never contains this image, and since $\pi_{4n-1}(\mathrm{Sp}(1))$ is torsion for $n\geq 2$ the above map can not do so either for $n\geq 2$.

(*) $\pi_{4n-1}(\mathrm{SO}(4n)) \cong \mathbb{Z}\times\mathbb{Z}$ follows WHEN $n\geq 4$ from the paper

Barratt, M. G.; Mahowald, M. E. The metastable homotopy of O(n). Bull. Amer. Math. Soc. 70 1964 775-760.

I think this is true in general. Indeed, it is true for $n=1$ where the above is not a contradiction because there $\pi_3(\mathrm{Sp}(1))\cong \mathbb{Z}$.

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