Let $f: \mathbb{R}^n \rightarrow \mathbb{C}$ be in $L^2\cap L^1,$ then the Fourier transform is in $L^2 \cap L^\infty.$


Does this imply that we can take common norms in the sense that we can estimate 


$$\sup\_{x\_i \in \mathbb{R}} \sqrt{\int\_{\mathbb{R}^{n-1}} \left\lvert \hat{f}(x\_1,...,x\_i ,...,x\_n) \right\rvert^2 dx\_1...dx\_{i-1}dx\_{i+1}...dx\_{n}}$$ or alternatively 
$$ \sqrt{\int\_{\mathbb{R}^{n-1}}\sup\_{x\_i \in \mathbb{R}} \left\lvert \hat{f}(x\_1,...,x\_i ,...,x\_n) \right\rvert^2 dx\_1...dx\_{i-1}dx\_{i+1}...dx\_{n}}?$$


Maybe this can even be bounded by mixed norms 


$$\int\_{\mathbb{R}} \sqrt{\int\_{\mathbb{R}^{n-1}} \left\lvert f(x\_1,...,x\_i ,...,x\_n) \right\rvert^2 dx\_1...dx\_{i-1}dx\_{i+1}...dx\_{n}}dx\_i$$ or


$$ \sqrt{\int\_{\mathbb{R}^{n-1}}\left(\int\_{\mathbb{R}} \left\lvert f(x\_1,...,x\_i ,...,x\_n) \right\rvert dx\_i\right)^2 dx\_1...dx\_{i-1}dx\_{i+1}...dx\_{n}}?$$


However, I do not see yet whether this is true or not.



Boundedness of different Fourier transforms
Yes, such bounds follow by considering partial Fourier transforms and then using the norm bounds you quoted. Let me take $n=2$ for ease of notation. Write $g(x,t)=\int f(x,k) e^{2\pi i kt}\, dk$. Then, for example (and as desired),
$$
\sup\_y \|\widehat{f}(x,y)\|\_{L^2(dx)} =\sup\_y \|g(x,y)\|\_{L^2(dx)}\le \left\| \int |f(x,k)|\, dk \right\|\_{L^2} \le \int\|f(x,k)\|\_{L^2(dx)}\,dk ,
$$
by the integral version of Minkowski's inequality for the final step.


