The most general, integral form of Faraday's Law is (see this physics.SE question: Faraday's law for a current loop being deformed)
\begin{align}
\int_{C_t} (\mathbf E+\mathbf v\times\mathbf B)\cdot d\boldsymbol \ell = - \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a
\end{align}
Where $C_t$ is some closed curve that can depend on time, $\Sigma_t$ is a surface with $C_t$ as its boundary, $\mathbf E$ and $\mathbf B$ are the electromagnetic fields as measured in some inertial frame, and $\mathbf v$ is the velocity of a point on the curve resulting from its time-dependence.
Now if we consider the situation you describe, then the $\mathbf v\times\mathbf B$ terms goes away if we choose a stationary loop $C=C_t$, and we get
\begin{align}
\int_{C}\mathbf E\cdot d\boldsymbol \ell = - \frac{d}{dt}\int_{\Sigma}\mathbf B\cdot d\mathbf a
\end{align}
Now you say that
all the magnetic field and hence flux is confined within the windings.
This is true. However you also say that
Therefore, for a wire making a loop surrounding the toroid and passing through the centre, the fields (electric and magnetic) at the wire is zero
This is not quite right. If the right hand side (the rate of change of the flux) is nonzero, then the line integral of the electric field around the loop must be nonzero.
\begin{align}
\int_{C}\mathbf E\cdot d\boldsymbol \ell \neq 0
\end{align}
In particular, this means that the electric field itself cannot vanish along the loop, otherwise we would have a contradiction. In other words, it may be the case that there is no magnetic field along the loop (at least at the initial instant before any current is generated), but there is an electric field along the loop, and this pushes charges around (if the loop is a conductor with charges in it). As a side note, once the charges start moving, they create their own magnetic field even in the absence of a magnetic field produced by the solenoid.
