I am using index notation here, since denoting traces in index notation is easier. Einstein summation convention assumed.
If $(M,g)$ is a Riemannian or pseudo-Riemannian manifold, and $$ R^\rho_{\sigma\mu\nu}=\partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+\Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}-\Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma} $$ is the Riemann curvature tensor, then the only independent trace of $R$ is the Ricci tensor $R_{\mu\nu}=R^\sigma_{\mu\sigma\nu}$, since the trace $R^\sigma_{\sigma\mu\nu}$ is zero.
If we are, on the other hand, given an arbitrary linear connection, it is necessarily a $\text{GL}(n,\mathbb{R})$-connection, and there is nothing specific to be said about the first two indices of the curvature tensor, so the tensor field $Q_{\mu\nu}=R^\sigma_{\sigma\mu\nu}$ is not necessarily zero.
What is there to be said about this tensor? What is its geometric meaning? What does it signify that for a Riemannian curvature tensor, this is zero?
I do realize that if $\nabla$ is an arbitrary $g$-compatible connection then, for and arbitrary frame $e_{a}$ (latin indices - frame indices, greek indices - coordinate indices) we have $$d^\nabla g_{ab}=0=dg_{ab}-\omega^c_{\ a}g_{cb}-\omega^c_{\ b}g_{ac},$$ so $dg_{ab}=\omega_{ba}+\omega_{ab}$, so if $e_a$ is an orthonormal frame then the connection forms are skew-symmetric, and then the curvature form $\Omega^a_{\ b}=d\omega^a_{\ b}+\omega^a_{\ c}\wedge\omega^c_{\ b}$ is also skew-symmetric. Moreover, since unlike $\omega$, $\Omega$ is gauge-covariant, this skew-symmetry is preserved even if calculated in a non-orthonormal frame.
Therefore, if $Q_{\mu\nu}$ does not vanish, then $\nabla$ cannot be metric-compatible for any metric I assume.
But I am curious about more info. Does the vanishing of $Q$ also imply that $\nabla$ is metric compatible for some metric? What else can be said about $Q$?
For any affine connection $\nabla$ on a smooth manifold, the curvature $R_{ab}{}^c{}_d$ may be uniquely decomposed as $$R_{ab}{}^c{}_d = C_{ab}{}^c{}_d + 2 \delta^c{}_{[a} {\mathsf P}_{b]d} + \beta_{ab} \delta^c{}_d \qquad (\ast)$$ for some totally tracefree $C$, called the projective Weyl tensor, and skew $\beta$; the tensor $\mathsf P$ is called the projective Schouten tensor. The First Bianchi Identity, $R_{[ab}{}^c{}_{d]} = 0$, implies that $-2 {\mathsf P}_{[ab]} = \beta_{ab}$.
Now, taking the trace over ${}^c{}_d$ gives $$Q_{ab} := R_{ab}{}^c{}_c = -2 {\mathsf P}_{[ab]} + n \beta_{ab} = (n + 1) \beta_{ab}.$$ Then, taking the trace of $(\ast)$ over ${}^c{}_a$ implies that $Q_{ab} = -2 R_{[ab]}$, so $Q$ is, up to a constant multiple, the skew part of the Ricci curvature of $\nabla$.
A more concrete geometric interpretation is this: Computing from $(\ast)$ gives that the curvature of the connection $\nabla$ induces on the anticanonical bundle $\Lambda^n TM$ is $Q$, or equivalently that the curvature of the connection induced on the canonical bundle $\Lambda^n T^*M$ is $-Q$. Thus, this bundle locally admits parallel sections, that is, $\nabla$ (locally) preserves a volume form on $M$, iff $Q = 0$, corresponding to the fact that (by definition) $Q = 0$ iff $R$ takes values in ${\frak sl}(TM)$. In particular, the Levi-Civita connection of any metric $g$ preserves the volume form of the restriction of that metric to any open orientable subspace (endowed with either choice of orientation), so $Q = 0$ for a Levi-Civita connection, or, like you say, for any metric connection.
Expanding the Second Bianchi Identity, $\nabla_{[e} R_{ab]}{}^c{}_d$, using $(\ast)$ implies that $dQ = 0$, so $Q$ defines a second cohomology class $[Q] \in H^2(M)$. If $\nabla$ is torsion-free, any connection projectively equivalent to $\nabla$, that is, sharing the same (unparameterized) geodesics as $\nabla$, has the form $$\hat\nabla_a \xi^b = \nabla_a \xi^b + \Upsilon_a \xi^b + \Upsilon_c \xi^c \delta^b{}_a$$ for some $\Upsilon \in \Gamma(T^*M)$ (and any choice of $\Upsilon$ gives projectively equivalent connections). The corresponding tensors $Q, \hat Q$ are related by $\hat Q = Q + 2 (n + 1) d\Upsilon$, and in particular, they differ by an exact form. Thus, the cohomology class $[Q] = [\hat Q]$ is actually an invariant of the projective structure---that is, the equivalence class of projective equivalent connections---that $\nabla$ defines. On the other hand, this transformation rule for $Q$ shows that locally $\nabla$ is projectively equivalent to one with $Q = 0$ (such connections are sometimes called special). So, in the setting of local projective differential geometry, we may as well just work with special connections, which enjoy the convenient feature that ${\mathsf P}_{ab}$ and $R_{ab}$ are symmetric.
This formulation can be found, by the way, in $\S$3 of the following reference:
It is not true that vanishing of $Q$ implies that $\nabla$ is a Levi-Civita connection, or even that it is projectively equivalent to one. Naively one should expect as much: Vanishing of $Q$ implies that the (local) holonomy group of $\nabla$ based at any point $x$ is contained in $\textrm{SL}(T_x M)$, but if $\nabla$ is the Levi-Civita connection of a metric $g$, the holonomy group must be contained in the much smaller group $\textrm{SO}(g_x)$.
A simple example is the connection $\nabla$ on $\Bbb R^3$ whose nonzero Christoffel symbols are specified (in the canonical coordinates $(x^a)$) by $\Gamma_{21}^3 = \Gamma_{31}^2 = x^2$. (The projective structure this connection defines is the so-called Egorov projective structure, which is interesting for other reasons, too.) The nonzero components of curvature are specified by $R_{23}{}^1{}_2 = -R_{32}{}^1{}_2 = -1$, so $\nabla$ is special, but $\S$2.3 of the below reference shows that it is not a Levi-Civita connection, nor even projectively equivalent to one.