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(I would like to ask this as a comment but alas I lack the required 50 rep to comment on other people's posts.)

Could you clarify a few things in your question? In particular I'm wondering how you're defining $f$ if $F$ isn't $L^1$; do you mean to take some sort of limit $f(x) = \lim_{R \to \infty} \int_{-R}^R F(y)^{ixy}dy$ (assuming $F$ is at least $L^1_{\text{loc}}$) or are you assuming $F$ is a tempered distribution and you're defining $f$ to be the distributional Fourier transform of $F$ (which agrees with the given formula when $F$ is $L^1$)? In particular, is your question (equivalent to) the following: if I have a tempered distribution $F$ and its distributional Fourier transform $f$ is a function, must this function be of polynomial growth? (In which case the Fourier transform is irrelevant.) If this is not your question, could you indicate what assumptions you're putting on the function/object $F$?

edited to add the following considerations:

if you only assume that $F \in \mathcal{S}^\prime$ or $F \in L^1_{\text{loc}}$, and you define $f$ by either (i) taking a limit as $R \to \infty$ in the limits of the integral, or (ii) by the distributional Fourier transform, then $f$ may be a function which is continuous almost everywhere but not of polynomial growth. For example, take $F$ to be the inverse Fourier transform of an $L^2$ function $f$ which is very 'spiky', e.g. $f = \sum_{n = 1}^\infty n^n \chi_{[n, n + n^{-3n}]}$. Note $||f||_{L^2}^2 = \sum_1^\infty n^{-n} < \infty$, and by construction $f$ is not of polynomial growth, though it is continuous almost everywhere. Clearly we recover $f$ by method (ii), and since $F\chi_{[-n,n]} \to F$ in $L^2$ as $n \to \infty$, we also recover $f$ both by method (i) (and moreover the limit as $R \to \infty$ of the integral exists almost everywhere by Carleson's theorem).

Edited to more accurately address the OPs (restated) concerns:

It appears that one cannot hope for a general result for $L^p$, analytic (even entire) $F$ if $p \geq 2$. Consider, for example $$ f = \sum_{n = 1}^\infty n^n \chi_{[n, n + e^{-n^2}]}. $$ (By mollifying the cutoffs one could also take $f \in C^\infty$, but let me not worry about that for now.) We have $$ ||f||_{L^p}^p = \sum_{n = 1}^\infty n^{pn}e^{-n^2} < \infty $$ for all $1 \leq p < \infty$ (but $f \notin L^\infty$).

For $\xi \in \mathbb{C}$, define $$ F(\xi) = \int_{\mathbb{R}}f(y)e^{-i\xi y}dy. $$ From the definition of $f$, one immediately sees this is well-defined (the integral is absolutely convergent), and the usual differentiation under the integral sign (again, readily justified from the explicit form of $f$) shows that $F$ is an entire function. By the Hausdorff-Young inequality, $F \in L^p(\mathbb{R})$ for all $2 \leq p \leq \infty$, though we can also say with certitude that $F \notin L^1(\mathbb{R})$, since its Fourier transform $f$ is not $L^\infty$. By Carleson's theorem, $f(x) = \lim_{R \to \infty} \int_{-R}^R F(y)e^{ixy}dy$ for $x$-a.e., and of course $f$ is not of polynomial growth.

(I would like to ask this as a comment but alas I lack the required 50 rep to comment on other people's posts.)

Could you clarify a few things in your question? In particular I'm wondering how you're defining $f$ if $F$ isn't $L^1$; do you mean to take some sort of limit $f(x) = \lim_{R \to \infty} \int_{-R}^R F(y)^{ixy}dy$ (assuming $F$ is at least $L^1_{\text{loc}}$) or are you assuming $F$ is a tempered distribution and you're defining $f$ to be the distributional Fourier transform of $F$ (which agrees with the given formula when $F$ is $L^1$)? In particular, is your question (equivalent to) the following: if I have a tempered distribution $F$ and its distributional Fourier transform $f$ is a function, must this function be of polynomial growth? (In which case the Fourier transform is irrelevant.) If this is not your question, could you indicate what assumptions you're putting on the function/object $F$?

edited to add the following (elementary) considerations:

if you only assume that $F \in \mathcal{S}^\prime$ or $F \in L^1_{\text{loc}}$, and you define $f$ by either (i) taking a limit as $R \to \infty$ in the limits of the integral, or (ii) by the distributional Fourier transform, then $f$ may be a function which is continuous almost everywhere but not of polynomial growth. For example, take $F$ to be the inverse Fourier transform of an $L^2$ function $f$ which is very 'spiky', e.g. $f = \sum_{n = 1}^\infty n^n \chi_{[n, n + n^{-3n}]}$. Note $||f||_{L^2}^2 = \sum_1^\infty n^{-n} < \infty$, and by construction $f$ is not of polynomial growth, though it is continuous almost everywhere. Since $F\chi_{[-n,n]} \to F$ in $L^2$ and $\mathcal{S}^\prime$ as $n \to \infty$, we recover $f$ both by method (i) and (ii) above.

(I would like to ask this as a comment but alas I lack the required 50 rep to comment on other people's posts.)

Could you clarify a few things in your question? In particular I'm wondering how you're defining $f$ if $F$ isn't $L^1$; do you mean to take some sort of limit $f(x) = \lim_{R \to \infty} \int_{-R}^R F(y)^{ixy}dy$ (assuming $F$ is at least $L^1_{\text{loc}}$) or are you assuming $F$ is a tempered distribution and you're defining $f$ to be the distributional Fourier transform of $F$ (which agrees with the given formula when $F$ is $L^1$)? In particular, is your question (equivalent to) the following: if I have a tempered distribution $F$ and its distributional Fourier transform $f$ is a function, must this function be of polynomial growth? (In which case the Fourier transform is irrelevant.) If this is not your question, could you indicate what assumptions you're putting on the function/object $F$?

edited to add the following considerations:

if you only assume that $F \in \mathcal{S}^\prime$ or $F \in L^1_{\text{loc}}$, and you define $f$ by either (i) taking a limit as $R \to \infty$ in the limits of the integral, or (ii) by the distributional Fourier transform, then $f$ may be a function which is continuous almost everywhere but not of polynomial growth. For example, take $F$ to be the inverse Fourier transform of an $L^2$ function $f$ which is very 'spiky', e.g. $f = \sum_{n = 1}^\infty n^n \chi_{[n, n + n^{-3n}]}$. Note $||f||_{L^2}^2 = \sum_1^\infty n^{-n} < \infty$, and by construction $f$ is not of polynomial growth, though it is continuous almost everywhere. Clearly we recover $f$ by method (ii), and since $F\chi_{[-n,n]} \to F$ in $L^2$ as $n \to \infty$, we also recover $f$ both by method (i) (and moreover the limit as $R \to \infty$ of the integral exists almost everywhere by Carleson's theorem).

(I would like to ask this as a comment but alas I lack the required 50 rep to comment on other people's posts.)

Could you clarify a few things in your question? In particular I'm wondering how you're defining $f$ if $F$ isn't $L^1$; do you mean to take some sort of limit $f(x) = \lim_{R \to \infty} \int_{-R}^R F(y)^{ixy}dy$ (assuming $F$ is at least $L^1_{\text{loc}}$) or are you assuming $F$ is a tempered distribution and you're defining $f$ to be the distributional Fourier transform of $F$ (which agrees with the given formula when $F$ is $L^1$)? In particular, is your question (equivalent to) the following: if I have a tempered distribution $F$ and its distributional Fourier transform $f$ is a function, must this function be of polynomial growth? (In which case the Fourier transform is irrelevant.) If this is not your question, could you indicate what assumptions you're putting on the function/object $F$?

(I would like to ask this as a comment but alas I lack the required 50 rep to comment on other people's posts.)

Could you clarify a few things in your question? In particular I'm wondering how you're defining $f$ if $F$ isn't $L^1$; do you mean to take some sort of limit $f(x) = \lim_{R \to \infty} \int_{-R}^R F(y)^{ixy}dy$ (assuming $F$ is at least $L^1_{\text{loc}}$) or are you assuming $F$ is a tempered distribution and you're defining $f$ to be the distributional Fourier transform of $F$ (which agrees with the given formula when $F$ is $L^1$)? In particular, is your question (equivalent to) the following: if I have a tempered distribution $F$ and its distributional Fourier transform $f$ is a function, must this function be of polynomial growth? (In which case the Fourier transform is irrelevant.) If this is not your question, could you indicate what assumptions you're putting on the function/object $F$?

edited to add the following (elementary) considerations:

if you only assume that $F \in \mathcal{S}^\prime$ or $F \in L^1_{\text{loc}}$, and you define $f$ by either (i) taking a limit as $R \to \infty$ in the limits of the integral, or (ii) by the distributional Fourier transform, then $f$ may be a function which is continuous almost everywhere but not of polynomial growth. For example, take $F$ to be the inverse Fourier transform of an $L^2$ function $f$ which is very 'spiky', e.g. $f = \sum_{n = 1}^\infty n^n \chi_{[n, n + n^{-3n}]}$. Note $||f||_{L^2}^2 = \sum_1^\infty n^{-n} < \infty$, and by construction $f$ is not of polynomial growth, though it is continuous almost everywhere. Since $F\chi_{[-n,n]} \to F$ in $L^2$ and $\mathcal{S}^\prime$ as $n \to \infty$, we recover $f$ both by method (i) and (ii) above.