We do not actually know whether a fifth root can be constructed from its radicand by neusis, nor do we know about quintisection of a given angle.
Both problems involve solutions of irreducible quintic equations, which would be impossible if neusis were merely equivalent to construction aided by conic sections or single-fold origami. But in 2002 Baragar showed that neusis is more powerful and demonstrated this extra power with a geometrical solution to the irreducible quintic equation
$$x^5 − 4x^4 + 2x^3 + 4x^2 + 2x − 6 = 0.$$
Benjamin and Snyder developed a method for solving sextic equations satisfying certain relationships using neusis and showed that one such soluble equation has the form
$$(x-u-2)(x^5+ux^4-4u^2x^3-3u^3x^2+3u^4x+u^5)=0$$
where the quintic factor gives $\dfrac xu=2\cos\left(\dfrac{k\pi}{11}\right)$. The scale factor $u$ solves a cubic equation $u^3+2u^2+2u+2=0$ and thus is marked ruler constructible. Thereby the regular hendecagon is neusis constructible.*
Neither of these constructions sheds any light on fifth root extraction or angle quintisection. Baragar's quintic is not radical-soluble. The equation for the hendecagonal cosines is soluble via complex fifth-root extraction, therefore angle quintisection, but Benjamin and Snyder's method obtains the polygon without using angle quintisection.
*We know now that the cubic equation for $u$ found by Benjamin and Snyder is not entirely accidental. The tribonacci ratio equation $t^3-t^2-t-1=0$, with roots closely related to Benjamin and Snyder's $u$ factor, is coupled to the original quintic equation through the quadratic Gauss sum of $11$th roots of unity. See https://mathoverflow.net/a/432460/86625 for details.