I have a set of equations to integrate something in time $t$. At each time step I compute a scalar field $\phi(t)$ and a potential $V(\phi)$. I should also control the conservation of energy with an equation:
$E=\frac{1}{2}\left(\dot\phi^2+V(\phi)\right)$
Or numerically:
$E=\frac{1}{2}\left(\left(\frac{\phi_{n}-\phi_{n-1}}{t_{n}-t_{n-1}}\right)^2+V(\phi)\right)$
My question is: is it $V(\phi_{n})$ or $V(\phi_{n-1})$ ?
It depends on how you want to solve it: via implicit or explicit methods.
The explicit method is certainly the easiest to code but constrains $dt$ for stability; the implicit method is a bit more difficult to code but has no such constraints on $dt$.
For example, if we consider a simple advection equation $$\frac{\partial w}{\partial t}+u\frac{\partial w}{\partial x}=0$$ Then the explicit method solves this as (using your notation, I think most people do $n\to n+1$, rather than $n-1\to n$) $$ \frac{w^{n}_j+w^{n-1}_j}{dt}=u\left[\frac{w_{j+1}^{n-1}-w_{j-1}^{n-1}}{2\cdot dx}\right] $$ which is more commonly written as $$ w^n_j=w^{n-1}_j+\chi u\left[w^{n-1}_{j+1}-w_{j-1}^{n-1}\right] $$ where $\chi=dt/2dx$. The implicit method solves this as $$ \frac{w^{n}_j+w^{n-1}_j}{dt}=u\left[\frac{w_{j+1}^{n}-w_{j-1}^{n}}{2\cdot dx}\right] $$ which is more commonly written as $$ -\chi uw_{j+1}^n+w^n_j-\chi uw^n_{j-1}=-w_j^{n-1} $$ which requires solving a tri-diagonal matrix (fortunately there is a relatively simple algorithm to solve this).
I generally would recommend trying the explicit method first and seeing if the time-stepping is good enough to be useful; if it's not, then it's time to use the implicit method.
I will just sum up the useful remarks made in the comments. Both approximations
$E=\frac{1}{2}\left(\left(\frac{\phi_{n}-\phi_{n-1}}{t_{n}-t_{n-1}}\right)^2+V(\phi_n)\right)$
and
$E=\frac{1}{2}\left(\left(\frac{\phi_{n}-\phi_{n-1}}{t_{n}-t_{n-1}}\right)^2+V(\phi_{n-1})\right)$
will give you first-order accurate approximation of the energy. If you wish to get higher than 1st order accuracy, you should use a more accurate finite difference approximation of $\dot\phi$. For instance, you could get second order with
$E=\frac{1}{2}\left(\left(\frac{\phi_{n+1/2}-\phi_{n-1/2}}{t_{n}-t_{n-1}}\right)^2+V(\phi_{n})\right)$.
The definition of the derivative is $$\dot{\phi}_n=\lim_{\delta t\rightarrow 0}\frac{\phi_{n+1}-\phi_n}{\delta t}$$ where $\delta t$ is $t_n-t_{n-1}$. Therefore, the derivative as you are taking it is calculated for the $n-1$-th step. Hence, I'd say you should go for $V(\phi_{n-1})$.