It is an interesting problem which is related to some recent work of mine. The reason for why I started to work on similar problems is because connections to a problem of Ramachandra on Dirichlet polynomials, connections to the nordic school of Hardy classes of Dirichlet series (Hedenmalm, Saksman, Seip, Olsen, Olofsson, Lindqvist and others), as well as universality questions for zeta-functions and their properties on the line Re(s)=1.
While my papers are not quite finished, I have put two early preprints on my homepage, On a problem of Ramachandra and approximation of functions by Dirichlet polynomials with bounded coefficients and On generalized Hardy classes of Dirichlet series. I have talked about some of these problems at analytic number theory conferences in India. Like in your paper I have considered Dirichlet series (it should be possible to obtain something like Theorem 2.1 in your paper by my method also, although I have not stated a direct analogue in my paper).
Now your problem in the question is rather easy for small $\omega$ so we will from now on assume that $\omega>1/2$. In fact if $\omega<1/2$, then $|f(0)|>1/2$ and $\int_0^K |\hat f(t)^2|dt \geq \min(1/10,K/10)$ (constants not chosen in an optimal way)
In my papers on Dirichlet series I have used a somewhat different method than you use in your paper, namely the Jensen inequality on the logarithmic integral in a half-plane. This method is applicable for the problem at hand. Lemma 7 in my paper ``On generalized Hardy classes of Dirichlet series'' can be used with $\sigma=0$ and $L(it)=\hat f(-t)$ and we obtain
$$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq \frac D \pi \int_{-\infty}^\infty \frac {\log^+ |\hat f (t)|} {D^2+t^2} dt - \log |\hat f(iD)|.
$$
For similar results see also Koosis - The logarithmic integral. (Remark Feb 16: The above inequality is an equality if the function is non-zero on a half plane. The inequality follows from Jensen's formula on a disc by mapping the half plane on the disc by the standard holomorphic bijection where $iD$ goes to $0$)
The reason why we can do this is that with the definition of the fourier-transform in your question it means that $ \hat f(z)$ will be a bounded analytic function in the half plane Im$(z) \geq 0$.
Now in this case we also have that $\log^+ |\hat f (t)|=0$ since $ |\hat f (t)| \leq 1$. Thus the inequality simplifies to
$$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq - \log |\hat f(iD)|.$$
It is not too difficult to see that for $\omega>1/2$
$$
|\hat f(i\omega)|= \left|\int_0^1 e^{i \phi(x)-\omega x} dx \right|>\frac {1} {10 \omega}.
$$
(The constant $10$ not chosen optimally). Thus we can choose $D=\omega$ and it is clear that
$$
\int_0^K \log^- |\hat f(t)| dt < \frac \pi {\omega} \left({\omega^2+K^2} \right) \frac {\omega} \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {\omega^2+t^2} dt
$$
From these estimates we see that
$$
\frac 1 K \int_0^K \log^- |\hat f(t)| dt< \frac {\pi(\omega^2+K^2)}{\omega K} \log (10 \omega).
$$
Now we can use the Jensen inequality
$$
\exp\left(\frac 1 K \int_0^K \log |\hat f(t)| dt\right)< \sqrt{\frac 1 K \int_0^K |\hat f(t)|^2 dt}
$$
We get the lower bound
$$
K \left(\frac 1 {10 \omega} \right)^{2\pi (\omega^2+K^2)/(K \omega)} \leq \int_0^K |\hat f(t)|^2 dt
$$
for $\omega>1/2$. If $c>2 \pi$ and $\omega/K$ is sufficiently large this gives a lower bound
$$\omega^{-c \omega/K} \leq \int_0^K |\hat f(t)|^2 dt$$
which is weaker than your expected $e^{-c \omega/K}$. At least we have an explicit lower bound.
Updated Feb 16: In the case where both $\omega$ and $K$ are large but still $\omega>K$ this can be improved by the following trick. Let $g$ be the convolution of $\hat f$
with a non negative test-function $\Phi(t/K)$, such that $\hat \Phi(0)>0$ where $\Phi$ has support on $[0,1/2]$ . Then use Jensen's inequalities on the function $g$ instead of $\hat f$ as above. The advantage with this is that it then follows that $|\hat g(iw)| \gg K/\omega$ and thus we can get the lower bound (by using Jensen's inequality w.r.t the L^1-norm instead of the L^2-norm.)
$$(\omega/K)^{-c \omega/K} \leq \frac 1 K \int_0^{K/2} |g(t)| dt$$
for some constant $c>0$. Since
$$ g(t)=\int_0^t \Phi((t-x)/K) \hat f(x) dx$$ it is clear by the triangle inequality that
$$\frac 1 K \int_0^{K/2} |g(t)| dt = \frac 1 K \int_0^{K/2} \left|\int_0^t \Phi((t-x)/K)\hat f(x) \right| dx \leq $$
$$\leq \frac 1 K \int_0^{K/2} |f(x)| dx \int_0^{K/2} |\Phi(x/K)| dx \leq c \int_0^{K/2} |\hat f(x)| dx $$
The inequality
$$K^{-1} (\omega/K)^{-c \omega/K} \leq \int_0^{K/2} |\hat f(t)|^2 dt$$
follows by the Cauchy-Schwarz inequality for some constant $c>0$.
This formula just use involves dimensionless quantity $\omega/K$ as expected. Since the function $E(K)$ is increasing in $K$ it gives the lower bound $E(K) > C_0 K^{-1}>0$ for $1 \leq \omega \leq K$ for some absolute constant $C_0$.
Yes indeed, for fractional $s$ it is related to Modified Bessel Functions of 2nd Kind (Macdonald Function). For $r=|\mathbf{x}|,\ a=|x_{n+1}|$ $$P(r,a)=\frac{a^{2s}}{(r^2+a^2)^{n/2+s}}$$ The Fourier Transform of an n-dimensional radial function $f(r)$ is a radial function in the trasformed space $\mathcal{F}(|\xi|)$. It is given by the following Hankel Transform. $$\mathcal{F}(|\xi|)=\frac{1}{|\xi|^{\frac{n}{2}-1}}\int_0^\infty r^{\frac{n}{2}-1}\left[r\,J_{\frac{n}{2}-1}(|\xi|\cdot r)\right]\cdot f(r)\,dr$$ See, for instance, Sec 11 pg. 63-65 in
Sneddon, Ian N., Fourier transforms, New York: McGraw-Hill Book Company, Inc. (1950). ZBL0038.26801. for a proof.
Also, Sneddon, Ian N., A note on some relations between Fourier and Hankel transforms, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 9, 799-806 (1961). ZBL0100.31501. Therefore $$\mathcal{F}_P(|\xi|,a)=\frac{a^{2s}}{|\xi|^{\frac{n}{2}-1}}\int_0^\infty J_{\frac{n}{2}-1}(|\xi|\cdot r)\cdot \frac{r^{\frac{n}{2}}\,dr}{(r^2+a^2)^{\frac{n}{2}+s}}$$ But integral is formula 6.565.4 in 7-th Edition of
Gradshteyn, I. S.; Ryzhik, I. M., Table of integrals, series, and products. Ed. by Alan Jeffrey. CD-ROM version 1. 0 for PC, MAC, and UNIX computers., San Diego, CA: Academic Press. (1996). ZBL0918.65001.
It reads (adapted to our notation), for $s>0$ $$\int_0^\infty J_{\frac{n}{2}-1}(|\xi|\cdot r)\cdot \frac{r^{\frac{n}{2}}\,dr}{(r^2+a^2)^{\frac{n}{2}+s}}=\left(\frac{|\xi|}{2}\right)^{\frac{n}{2}+s-1}\cdot\frac{K_s(a\,|\xi|)}{a^s\,\Gamma(\frac{n}{2}+s)}$$ Being $K_s(\cdot)$ the Modified Bessel Function of 2nd Kind. Therefore, $$\mathcal{F}_P(|\xi|,a)=\frac{(a\,|\xi|)^sK_s(a\,|\xi|)}{2^{\frac{n}{2}+s-1}\Gamma(\frac{n}{2}+s)}$$
Since $K_\frac{1}{2}(z)=\sqrt{\frac{\pi}{2z}}\cdot e^{-z}$, for $s=\frac{1}{2}$ we get $$\mathcal{F}_P(|\xi|,a)=\frac{\sqrt{\pi}}{2^\frac{n}{2}\Gamma\left(\frac{n+1}{2}\right)}\cdot e^{-a\,|\xi|}$$ Note: Dimension constant and $2\pi$ in exponent are matter of Fourier Transform's normalization and convention used respectively. By 'Rosetta Table' Formulae 504 at the end of this Wiki page this can be fitted accordingly.
Using question convention -FT unitary, ordinary frequency, $\mathcal{F}^*$- and notation we get,
$$P(|x|,x_{n+1})=\frac{2^{s-\frac{1}{2}}\Gamma\left(\frac{n}{2}+s\right)}{\pi^{\frac{n+1}{2}}}\cdot\frac{x_{n+1}^{2s}}{(|x|^2+|x_{n+1}|^2)^{n/2+s}}$$ producing this amazingly simple expression $$\mathcal{F}_P^*(|\xi|,x_{n+1})=\sqrt{\frac{2}{\pi}}\cdot \left(2\pi\,x_{n+1}\,|\xi|\right)^s\cdot K_s\left(2\pi\, x_{n+1}\,|\xi|\right)$$
This gives $\exp(-2\pi |\xi| x_{n+1})$ for $s=1/2$.
Being so simple, this result should be added as new Formulae 505 in Wiki's Table Fourier Transforms of n-dimensional functions.
I leave this for further reading, a nice article on Fourier Transforms of Fractional Poison Kernels and Fractional Laplacian (See Section 2.8) at
Kwaśnicki, Mateusz, Ten equivalent definitions of the fractional Laplace operator, Fract. Calc. Appl. Anal. 20, No. 1, 7-51 (2017). ZBL1375.47038.
Best Answer
Yes. Let $\phi:(E,+)\rightarrow S^1$ be a continuous homomorphism. For each $x\in E$, the map $\mathbb R\rightarrow S^1; t\mapsto \phi(tx)$ is continuous and a homomorphism, so there is some $\mu(x)\in\mathbb R$ with $$\phi(tx) = \exp(i t \mu(x)) \qquad (x\in E,t\in\mathbb R).$$ Then, for $x,y\in E$ and $\lambda\in\mathbb R$, $$ \exp(it\mu(\lambda x+y)) = \phi(t(\lambda x+y)) = \phi(t\lambda x)\phi(ty) = \exp\big(it(\lambda\mu(x)+\mu(y)\big). $$ Letting $t\rightarrow 0$, we conclude that $$\mu(\lambda x+y) = \lambda\mu(x) + \mu(y).$$ So $\mu$ is a linear map. If $x_n\rightarrow 0$ and $\mu(x_n)\rightarrow\mu$ then $$ \exp(it\mu) = \lim_n \exp(it\mu(x_n)) = \lim_n \phi(tx_n) = \phi(tx) = \exp(it\mu(x)), $$ for all $t$, so again, $\mu=\mu(x)$, and we conclude that $\mu$ is continuous.
So $\mu\in E^*$ and $\phi(x) = e^{i\mu(x)}$ as you want.