Since we are allowed to switch crossing points, it suffices to consider a knot projection (a knot diagram without over/undercrossing information, or equivalently, an immersed curve with finitely many self-crossings) on the plane. It is well known that there exists a sequence of (flat) Reidemeister moves which transforms any such curve to a simple closed curve such that the number of crossings is non-increasing, just as Neil mentioned. This result also holds even if one replaces the plane (or $S^2$) with oriented closed surfaces with higher genera [Hass J, Scott P (1994) Topology].

$T^3$ cannot be obtained by Dehn surgery on the unknot in $S^3$, since $T^3$ has Heegaard genus 3, which is not a lens space.

If the fundamental group has $n$ generators, then the bridge number (=crookedness) is at least $n$, which implies the minimal curvature is at least $2\pi n$ (Theorem 4.7 in Milnor's paper).

@Charles: $=\frac{1}{8}\times3^{3n}(3^n+1)^3(3^n+2)^3(3^{2n-1}+2\times3^{n-1}+1)^3$ $=[\frac{1}{2}\times 3^n(3^n+1)(3^n+2)(3^{2n-1}+2\times3^{n-1}+1)]^3$ The number is $y-x+1=(3^n+1)^3$ Therefore $\sum\limits_{i=0}^{(3^n+1)^3-1}(\frac{1}{2}\times3^{n-1}(3^{3n}+3^{2n}-5\times3^n-9)+i)^3=\frac{1}{2}\times3^n(3^n+1)(3^n+2)(3^{2n-1}+2\times3^{n-1}+1)$

Here is a translation of the picture above: PS: Absolute (Fermat) pseudoprimes that can be written as the product of three (prime?) factors can be completely classified. For example, the two simplest formulas which contain the quadratic term are 1. $(6n+1)(18n+1)(54n^2+12n+1)$; 2. $(1764n-139)(2268n-179)(1000188n^2-157752n+6221))$. 7 is a magic number. A deformation of the Fermat's little theorem (with the same base) $\frac{(\frac{N^{p_1}-N}{N}-\frac{N^{p_1-p_2+1}-N}{p_1-p_2})(p_1-p_2)}{p_2}(N, p_1, p_2$

If each letter is chosen randomly from the braid generators, the closure may not be a knot. If you do not insist that the closure must be a knot. A result in [Jiming Ma, THE CLOSURE OF A RANDOM BRAIDIS A HYPERBOLIC LINK, PAMS, 142(2), 2014, 695-701] says that the probability that the closure is a hyperbolic link converges to 1 when the length $l\rightarrow\infty$.

Here is a simple comment: each two-component link can be realized in this way. In fact, each knot in $S^3$ bounds an immersed disk with some clasp singularities. You can choose one clasp disk for each component respectively, and perturb them into general position. By taking a band sum of them you will obtain the desired immersed disk. For links with components more than 2 this is also correct.