First, note that you only get an induced $F^* $ map on cohomology if you fix a map $F^* \mathcal F \to \mathcal F$. Note that you get this for free if $\mathcal F$ is defined over $\mathbb F_p$.

The Frobenius action on the cohomology group of a sheaf is the main object of study in etale cohomology over a finite field. Entire books have been written about the action of Frobenius on the etale cohomology of certain sorts of sheaves. Thus, you shouldn't expect to easily escape this difficulty.

The reason that the induced map $\mathcal F \to \bar{\mathbb Q}_l$ doesn't work is that the map fits into a short exact sequence, giving a long exact sequence on cohomology and unless you have a good understanding of the kernel of that sequence there is no hope of computing the cohomology groups of $\mathcal F$. If $\mathcal F$ is very close to $\bar{\mathbb Q}_l$, and say the kernel is supported on a subvariety of smaller dimension, this is the right approach.

The reason that the other thing doesn't work is that it's not in general true. To see this, suppose $\mathcal F$ is something nice like the extension by $0$ of a lisse sheaf $\mathcal G$ on an open set $U$. Then the compact cohomology of $\mathcal F$ is the compact cohomology of $\mathcal G$, which is Poincare dual to the regular cohomology of $G^\vee$. So $H^{2d}_c(\mathcal F) = H^0(G^\vee)^{\vee}(-d)$.

Since it is often easy to compute the action of $F$ on the global sections of $G^v$, this is usually the best approach, as long as the sheaf is actually the extension by zero of a lisse sheaf. But note that by choosing $G$s dual to sheaves with different groups of global sections, you can get many different sorts of cohomology groups with different $Frob_p$ actions which are not the same as the $Frob_p$ action on $H^{2d}_c(\bar{\mathbb Q}_l)$.

First, note that you only get an induced $F^* $ map on cohomology if you fix a map $F^* \mathcal F \to \mathcal F$. Note that you get this for free if $\mathcal F$ is defined over $\mathbb F_p$.

The Frobenius action on the cohomology group of a sheaf is the main object of study in etale cohomology over a finite field. Entire books have been written about the action of Frobenius on the etale cohomology of certain sorts of sheaves. Thus, you shouldn't expect to easily escape this difficulty.

The reason that the induced map $\mathcal F \to \bar{\mathbb Q}_l$ doesn't work is that the map fits into a short exact sequence, giving a long exact sequence on cohomology and unless you have a good understanding of the kernel of that sequence there is no hope of computing the cohomology groups of $\mathcal F$. If $\mathcal F$ is very close to $\bar{\mathbb Q}_l$, and say the kernel is supported on a subvariety of smaller dimension, this is the right approach.

The reason that the other thing doesn't work is that it's not in general true. To see this, suppose $\mathcal F$ is something nice like the extension by $0$ of a lisse sheaf $\mathcal G$ on an open set $U$. Then the compact cohomology of $\mathcal F$ is the compact cohomology of $\mathcal G$, which is Poincare dual to the regular cohomology of $\mathcal G^\vee$. So $H^{2d}_c(\mathcal F) = H^0(\mathcal G^\vee)^{\vee}(-d)$.

Since it is often easy to compute the action of $F$ on the global sections of $\mathcal G^v$, this is usually the best approach, as long as the sheaf is actually the extension by zero of a lisse sheaf. But note that by choosing $\mathcal G$s dual to sheaves with different groups of global sections, you can get many different sorts of cohomology groups with different $Frob_p$ actions which are not the same as the $Frob_p$ action on $H^{2d}_c(\bar{\mathbb Q}_l)$.

In fact, the "typical" thing, in a lot of situations, is for $\mathcal G^\vee$ to have no global sections at all. This gives you a nice answer, but perhaps not the one that you were looking for. To say more I'd have to know more about your sheaf.