Here is a somewhat different way from Johan's of looking at this
problem. At each stage of the walk, choose a number $x$
uniformly from $[0,1]$ and then walk either a distance $x$ to the
right or $1-x$ to the left. This does not affect the probability
of becoming negative since there is still a uniform probability
of taking a step whose length belongs to the interval
$[-1,1]$. However, it does have the property that after taking
$n$ steps and choosing $0\leq x\leq 1$, the two possible
locations following the next step are the same modulo 1. Hence
the walk can be described as follows. Choose $n$ numbers
$0\lt x_1\lt \cdots\lt x_n\lt 1$, a sequence
$\epsilon=(\epsilon_1,\dots,\epsilon_n)$ of signs $\pm 1$, and a
permutation $w$ of $1,2,\dots,n$. Let the location be $y_k$ after
the $k$th step. If $\epsilon_k=1$ then step to the least real
number $y_{k+1}\equiv x_{w(k+1)}$ (mod 1), $y_{k+1}>y_k$. If
$\epsilon_k=-1$ then step to the greatest real number
$y_{k+1}\equiv x_{w(k+1)}$ (mod 1), $y_{k+1}\lt y_k$. But the
question of whether any $y_k$ is negative depends only on $\epsilon$ and
$w$, not the choice of $x_1,\dots,x_n$. There are $2^n n!$ ways
to choose $\epsilon$ and $w$. Is there a simple combinatorial
argument that the number of choices such that each $y_k>0$ is
$(2n-1)!!=1\cdot 3\cdot 5\cdots (2n-1)$? Then the probability of
success is $(2n-1)!!/2^nn! = (2n)!/4^nn!^2$.
Here is a reformulation of the combinatorial result that needs a
simple direct proof.
Let $f(n)$ be the number of pairs $(a_1a_2\cdots a_n,
b_1b_2\cdots b_{n-1})$ such that (a) $a_1 a_2\cdots a_n$ is a
permutation of $1,2,\dots, n$, (b) $b_i=0$ or $1$ if $a_i\lt
a_{i+1}$, (c) $b_i=0$ or $-1$ if $a_i>a_{i+1}$, and (d) $b_1
+b_2+\cdots+b_j\geq 0$ for all $1\leq j\leq n-1$. Then
$f(n)=(2n-1)!!$.
Update. The combinatorial result is proved bijectively by
O. Bernardi, B. Duplantier, and P. Nadeau in Séminaire
Lotharingien de Combinatoire, B63e (2010). In their citation [1]
they use this result for the same purpose as above, i.e., to
compute the probability $P_n$ (though they state the result a
little differently).
Second update. The method above can be applied to the $[l,r]$
generalization mentioned by Lwins in his comment. By rescaling we
may assume $l=-1$. If we are at $y$ sometime during the walk, choose
a number $x$ uniformly from $[0,1]$. With probability 1/2 step from
$y$ to $y+\frac{r-1}{2}+\frac{r+1}{2}x$. With probability 1/2 step
from $y$ to $y-1-\frac{r+1}{2}x$. This gives a uniform probability
of stepping from $y$ to a point in the interval $[y-1,y+r]$. It has
the property that once $x$ is chosen, the value of $y$ is determined
modulo $\frac{r+1}{2}$. Thus the walk can be described as follows:
pick uniformly and independently $0\lt x_1\lt \cdots\lt x_n \lt
\frac{r+1}{2}$,
pick a permutation $w$ uniformly from the symmetric group
$S_n$, and a sequence $\epsilon=(\epsilon_1,\dots,\epsilon_n)$ of
independently distributed signs, with a probability of
$\frac{r}{r+1}$ for a plus sign and $\frac{1}{r+1}$ for a minus sign.
Go through the same procedure as above, working mod
$\frac{r+1}{2}$ instead of mod 1. Again a proper walk (i.e., one
which never becomes negative) depends only on $w$ and $\epsilon$,
and we get the following result:
Theorem. The probability $P_n(r)$ that the walk is proper is
given by
$$ P_n(r) = \frac{1}{(1+r)^nn!}\sum r^{1+f(w,\beta)}, $$
summed over all pairs $w=a_1a_2\cdots a_n\in S_n$ and
$\beta=(b_1,\dots, b_{n-1})\in \lbrace 0,\pm 1\rbrace^n$ satisfying
the conditions (b) and (c) above, where $f(w,\beta)$ is the number
of integers $1\leq i\leq n-1$ for which either $a_i\lt a_j$ and
$b_i=0$, or $a_i\gt a_j$ and $b_i=1$.
For instance, $P_2(r)= (r+2r^2)/2(r+1)^2$ and $P_3(r)
=(r+8r^2+6r^3)/6(r+1)^3$. I conjecture that the numerator $N_n(r)$
of $P_n(r)$ is just the polynomial $\sum B_{n,i}r^i$ defined by
equation (4) of
http://math.mit.edu/~rstan/pubs/pubfiles/29.pdf. This paper gives
some additional information about the polynomials $\sum
B_{n,i}r^i$. Much additional information can be found in the
literature on Stirling permutations, e.g., Bona proves in
http://wenku.baidu.com/view/dfa70012cc7931b765ce15e4.html that all
zeros of this polynomial are real.
Third update. Alas, the conjecture in my second update is
false. Unless there is an error in my code, the sequence of
coefficients of $N_n(r)$ for $2\leq n\leq 7$ are $(1,2)$,
$(1,8,6)$, $(1,25,55,24)$, $(1,69,361,394,120)$,
$(1,176,1999,4416,3083,720)$,
$(1,426,9836,41019,52193,26620,5040)$. It is easy to see why
the leading coefficient of $N_n(r)$ is $n!$.
