Fibonorial of a fractional or complex argument
(Main Result: I have a formula for what you want, its practical and it works!)
Since you ask multiple questions, I suppose what follows isn't a complete answer. I learned this trick the other day, so I'm more than happy to share it here,
If you want an integral that can generalize the recurrence,
$$(1) \quad \mathfrak F(0)=1,\quad \mathfrak F(n+1)=\mathfrak F(n)\times F(n+1).$$
You can do so by assuming the formula is of a certain form and that $k(t,n)$ satisfies a delay differential equation.
$$(2) \quad \mathfrak F(n)=\int_0^{\infty} e^{-t} \cdot k(t,n) \ dt$$
We can then demand that the eigenvalue for the differential operator is equal to $F(n+1)$.
$$(3) \quad \partial_t [k(t,n+1)]=F(n+1) \cdot k(t,n)$$
It then follows that applying integration by parts to $(2)$ yields,
$$\mathfrak F(n+1)=\int_0^{\infty} e^{-t} \cdot k(t,n+1) \ dt={F(n+1)} \cdot \int_0^{\infty} e^{-t} \cdot k(t,n) \ dt={F(n+1)} \cdot \mathfrak F(n)$$
$$\Rightarrow \mathfrak F(n+1)=\mathfrak F(n)\cdot F(n+1).$$
To obtain an explicit form for the integration kernel, we must solve $(3)$, which is classified as a delay differential equation. Whether or not an explicit solution exists is uncertain without further investigation. However, the kernel can certainly be found numerically.
However, we can certainly define the kernel that satisfies $(3)$ to be the Fibonacci kernel.
Tentative (Approximate) Solution
I didn't post this originally because its a bit cheap for a solution. Effectively, this solution indexes the new factorial differently. Set,
$$k(t,n)=t^{F(n)}$$
And also we have,
$$F(z)=(\phi^z-\cos(\pi \cdot z) \cdot \phi^{-z})/\sqrt{5}$$
Then we have,
$$\mathfrak F(n+1)=\mathfrak F(F^{-1}(F(n+1)-1)) \cdot F(n+1)$$
instead of,
$$\mathfrak F(n+1)=\mathfrak F(n) \cdot F(n+1)$$
Asymptotics and a Formula
I've switched notation since its the reason I came up with this. According to Wolfram Alpha, we have this relation for the asymptotics of the fibonorial,
$$(4) \quad n!_F \sim C \cdot \cfrac{\phi^{n \cdot (n+1)/2}}{5^{n/2}}$$
Given this, lets hope we can find a function $g(n,t)$ such that,
$$(5) \quad \lim_{n \rightarrow \infty} \cfrac{n!_F \cdot g(n,t)}{(n+t)!_F}=1$$
I find that, and I leave you to verify, that,
$$(6) \quad g(n,t)=5^{-t/2} \cdot \phi^{(t^2/2+t/2)} \cdot \phi^{n \cdot t}$$
Now that we know what $g(n,t)$ is, multiply by both sides by $t!_F$ and simplify, assuming $t$ is a positive integer. We get,
$$(7) \quad t!_F=\lim_{n \rightarrow \infty} \cfrac{n!_F \cdot g(n,t)}{F(1+t) \cdot F(2+t) \ ...\ F(n+t)}$$
Now we formally extend $t$ from $N \gt 0$ to $t \in C$. Using this, I can construct (perhaps the first?!) table of "F-Gamma" values using $n!_F=\mathfrak F(n+1)$,
$$ \begin{array}{c|lcr} n & \mathfrak F(n) & \ln(\mathfrak F(n))\\ \hline 1 & 1 & 0\\ 1.1 & 0.966975 & -0.033582\\ 1.2 & 0.951630 & -0.049579\\ 1.3 & 0.951521 & -0.049692\\ 1.4 & 0.963404 & -0.037283\\ 1.5 & 0.982610 & -0.017543\\ 1.6 & 1.003127 & 0.003122\\ 1.7 & 1.018567 & 0.018396\\ 1.8 & 1.023911 & 0.023630\\ 1.9 & 1.017193 & 0.017048\\ 2.0 & 1 & 0\\ \end{array} $$
Here's a graph (I didn't say it'd be pretty ;))
Check the recurrence, defined in $(1)$. It works!!!
Extending what Zach466920 said:
$$\partial_t[k(t,n)]=F(n+1)\cdot k(t,n+1)\tag1$$
We can develop a power series solution for this. Let $$k(t,n)=\sum_{m=0}^\infty \kappa(n, m)\ t^m$$
We equation $(1)$ as: $$\sum_{m=1}^\infty m\ \kappa(n, m)\ t^{m-1} =\sum_{m=0}^\infty(m+1)\ \kappa (n, m) \ t^m =F(n+1)\cdot\sum_{m=0}^\infty \kappa(n+1, m)\ t^m$$ Which gives us the recurrence relation: $$(m+1)\ \kappa(n, m)=F(n+1)\ \kappa (n+1,m)$$ I have no idea how to solve this recurrence equation or even if it is solvable. However if it is you should be able to get a series out of it. It then becomes a problem of solving that series until you can finally find a closed form for $k$.
With help from @Semiclassical we have the solution for $\kappa(n, m)$ as $$\kappa(n,m)=\left(\prod^n_{m=0}\frac{m+1}{F(n+1)}\right)\kappa(0, m)=\frac{(m+1)^{n+1}}{\mathfrak{F}(n+1)}\kappa(0, m)$$ The question is now finding $k(t, n)$ from this relation.
We have that $k(0, n) = 0$ or else there will be trouble with the convergence of $\mathfrak F$. This means that $$\mathfrak F(n) = \sum_{m=1}^\infty \kappa(n, m) \int_0^\infty e^{-t} t^m dt=\sum_{m=1}^\infty \kappa(n, m) \Gamma(m)$$ Consequently $$\sum_{m=1}^\infty \kappa(0, m) \Gamma(m)=1\tag 2$$ Which means that the seeding terms of $\kappa(n, m)$ seem to shrink faster than $\Gamma(m)$ grows as $m\to \infty$ and $n=0$.
Once a solution can be found for whatever $\kappa(0, m)$ is then a series for $k(t, n)$ can be established. I did some experimentation to see if $\kappa(0, m)$ could be guessed by letting $$\kappa(0, m)=\frac1{2^m\Gamma(m)}$$ as to satisfy condition $(2)$. This however lead to a value of $\mathfrak F(0)=\sqrt{3}$. This is clearly wrong.
If we let $\kappa(0, m)=\frac{a_m}{\Gamma(m)}$ we have $$\mathfrak F(n) = \sum_{m=1}^\infty \frac{(m+1)^{n+1}}{\mathfrak (n+1)}a_{m}\\ \implies \mathfrak F(n)^2F(n+1)=\sum_{m=1}^\infty (m+1)^{n+1} a_m$$ If we let $n=0$ we have $$1=\mathfrak F(0)=\sum_{m=1}^\infty a_m(m+1)=\sum_{m=1}^\infty m a_m + \underbrace{\sum_{m=1}^\infty a_m}_{=1\text{ by }(2)}$$ Therefore: $$\sum_{m=1}^\infty ma_m = 0$$ No idea what this means and I am unsure whether further manipulation will actually get anywhere. I feel quite lost carrying these last few out. I am open to ideas on how to determine $a_m$.