Consider a bivariate zero-mean VAR(1) model
\begin{aligned}
y_{1,t} &= a_{1,1} y_{1,t-1} a_{1,2} y_{2,t-1} + u_{1,t}, \\
y_{2,t} &= a_{2,1} y_{1,t-1} a_{2,2} y_{2,t-1} + u_{2,t}
\end{aligned}
or in matrix notation, $\mathbb{y}_t=A \mathbb{y}_{t-1} + \mathbb{u}_t$.
Both $y_{1,t}$ and $y_{2,t}$ are dependent variables, each in its own equation, so they are explained by the model. Thus, they are endogenous. The lags, on the other hand, might seem to be exogenous, as they are not explained by the model... or are they?
The model applies for every $t$. E.g. for $t=6$,
\begin{aligned}
y_{1,6} &= a_{1,1} y_{1,5} a_{1,2} y_{2,5} + u_{1,6}, \\
y_{2,6} &= a_{2,1} y_{1,5} a_{2,2} y_{2,5} + u_{2,6}
\end{aligned}
and for $t=5$,
\begin{aligned}
y_{1,5} &= a_{1,1} y_{1,4} a_{1,2} y_{2,4} + u_{1,5}, \\
y_{2,5} &= a_{2,1} y_{1,4} a_{2,2} y_{2,4} + u_{2,5}.
\end{aligned}
Thus, if we decrease $t$ by 1, the seemingly exogenous variables $y_{1,5}$ and $y_{2,5}$ turn out to be endogenous, as the model explains them, too. Therefore, all the $y$ variables can be considered endogenous in this model.
(Also, for a specific time period such as $t=6$, $y_{1,5}$ and $y_{2,5}$ are predetermined. This is a useful feature for deriving some desirable properties of least squares or other estimators of the models parameters.)
You can now generalized from the bivariate zero-mean VAR(1) to a $k$-variate VAR($p$) model with nonzero mean, and the same logic goes through.