I was wondering... Is every symplectic connection $\nabla$ on some symplectic manifold $(M,ω)$ the LeviCivita connection of some metric $g$ on $M$? What about the local statement?
For any Riemannian metric $g$ on a symplectic manifold $(M, \omega)$ there exists (canonical) almost complex structure $J$ making these threetensors compatible (any one can be defined by other 2). compatible with $\omega$, which means that $\omega(\_, J\_)$ defines a Riemannian metric $\widetilde{g}$. For such a compatible triple and the LeviCivita connection $\widetilde{\nabla}$ the equation $\widetilde{\nabla} \omega = 0$ is equivalent to $J$ being integrable thus making $(M,\widetilde{g},\omega, J)$ into Kähler manifold. There is plenty of symplectic manifolds which are not Kähler. So if there is metric $g$ for a nonKähler Fedosov manifold $(M, \omega, \nabla)$ whose LeviCivita connection is $\nabla$ then it cannot be compatible with $\omega.$

$\begingroup$ Oh, the question was so dumb! The answer for the local query is affirmative and is an immediate consequence of Darboux's theorem. Thank you, Vít Tuček! $\endgroup$ – Valentino Nov 26 '18 at 23:39

$\begingroup$ Havin an easy answer doesn't make a question dumb. I think it's a good question and it's answer requires some nontrivial theorems. $\endgroup$ – Vít Tuček Nov 27 '18 at 11:32

$\begingroup$ Wait... I was so excited that I didn't realize a lack of understanding of mine: why is it true that any Riemannian metric is the metric of some compatible triple? $\endgroup$ – Valentino Nov 27 '18 at 13:40

$\begingroup$ See here: math.stackexchange.com/questions/2084282/… $\endgroup$ – Paul Bryan Nov 30 '18 at 4:55

$\begingroup$ @Valentino I don't know! The answer is wrong as it stands as the metric $g$ is in general NOT the compatible metric for $J$ and $\omega$. I will revise the answer. $\endgroup$ – Vít Tuček Nov 30 '18 at 10:45
I think I've got the answer, and it is "no", at least for the global question.
When I started to try to understand the problem, I realized that to get some structure that is invariant by the connection we can (have to) fix the structure at some point and use parallel transport through picewise smooth paths to extend it. The resulting structure will be welldefined iff the initially fixed structure is invariant by the holonomy group at the point. The idea is then to construct some connection whose holonomy group at some point preserves some symplectic structure on the tangent space of the point but does not preserve any metric on such tanget space. To get the second feature, we may try to construct the connection in such a way that its holonomy group is unbounded. After some tries, I got the following:
On $\mathbb{R}^2$, we consider the connection $\nabla$ given by \begin{align*} \nabla_{\frac{\partial}{\partial x}}\frac{\partial}{\partial x} & = y\frac{\partial}{\partial x} & \nabla_{\frac{\partial}{\partial x}}\frac{\partial}{\partial y} & = 0 \\ \nabla_{\frac{\partial}{\partial y}}\frac{\partial}{\partial y} & = x\frac{\partial}{\partial y} & \nabla_{\frac{\partial}{\partial y}}\frac{\partial}{\partial x} & = 0. \end{align*}
The curvature $R$ of $\nabla$ at $(0,0)$ is given by $R(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \in GL(\mathbb{R}^2) = GL(T_{(0,0)}\mathbb{R}^2)$, and all its covariant derivatives are of the form $\begin{bmatrix} c & 0 \\ 0 & c \end{bmatrix}$ at such point. Because of that and of the connectedness of $Hol(\nabla, (0,0))$ (simplyconnectedness of $\mathbb{R}^2$), AmbroseSinger's theorem implies that the canonical symplectic form of $T_{(0,0)}\mathbb{R}^2 = \mathbb{R}^2$ is invariant by $Hol(\nabla, (0,0))$. On the other hand, $Hol(\nabla, (0,0))$ contains the exponential of $tR(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$ for every $t \in \mathbb{R}$, and is therefore an unbounded subset of $GL(T_{(0,0)}\mathbb{R}^2)$. As such, it doesn't fix any metric on $T_{(0,0)}\mathbb{R}^2$.
$ \textit{edit:}$ using LeviCivita's formula, it is easy to conclude that the connection $\nabla$ defined above is not the LeviCivita connection of any metric, even locally. With that in mind, Darboux's theorem shows us that for every symplectic manifold there exist some open subset $U$ of the manifold and some symplectic connection $\nabla$ defined on $U$ such that $\nabla$ is not the LeviCivita connection of any metric on $U$.
By using partitions of unity, we conclude, then, that $\textbf{every symplectic manifold admits}$ $\textbf{a symplectic connection that is not a}$ $\textbf{LeviCivita connection.}$