There is a general tactics for this. Recall that the three-term recurrence for the Legendre polynomials comes from a differential equation for their generating function.
The general background:
Generally, if a power series $f=\sum\limits_{i=0}^{\infty} p_it^i$ (where the coefficients $p_i$ belong to some field $K$ (for example, we can have $K=\mathbb Q$ or $K=\mathbb C\left(x\right)$ or whatever)) satisfies a differential equation $\sum\limits_{k=0}^n a_kf^{(k)}=0$ with coefficients $a_k\in K\left[t\right]$ (that's right, they can be polynomials in $t$), then we get a recurrence equation for the coefficients $p_i$ of $f$ in the following way:
Since $f=\sum\limits_{i=0}^{\infty} p_it^i$, we have $f^{(k)}=\sum\limits_{i=0}^{\infty} \left(i+k\right)\left(i+k-1\right)...\left(i+1\right)p_{i+k}t^i$. Hence, the equation $\sum\limits_{k=0}^n a_kf^{(k)}=0$ becomes $\sum\limits_{k=0}^n a_k\sum\limits_{i=0}^{\infty} \left(i+k\right)\left(i+k-1\right)...\left(i+1\right)p_{i+k}t^i=0$. Now write each $a_k$ as a polynomial in $t$, and expand the product on the left hand side. You get an equation of the form $\sum\limits_{u=0}^{\infty}\left(\text{some }K\text{-linear combination of }p_{u-m},p_{u-m+1},...,p_{u+n}\right)t^u=0$, where $m$ is $\max\left\lbrace\deg a_0,\deg a_1,...,\deg a_n\right\rbrace$. Now, equating coefficients, we get a recurrence equation for the $p_i$. Sometimes it may "degenerate", i. e. for some $i$ the coefficient before $p_i$ may be zero, but this shouldn't happen too often.
[Note that this recurrence equation has finitely many terms, since $a_k$ are polynomials in $t$. Had we allowed $a_k$ to be arbitrary power series in $t$, we would have obtained an infinite recurrence.]
Applications to the question:
The generating function for the Legendre polynomials is $\dfrac{1}{\sqrt{1-2xt+t^2}}$; this means that $\dfrac{1}{\sqrt{1-2xt+t^2}}=\sum\limits_{n=0}^{\infty}P_nt^n$ as a power series in the indeterminate $t$ over $K=\mathbb Q\left[x\right]$, where I write $P_n$ for $P_n\left(x\right)$.
Now you are interested in finding a recurrence relation for the $na_n=nP_n-\left(n-1\right)P_{n-2}$ (here I have multiplied by $n$ to get rid of the ugly fraction). So you are looking for a differential equation - with coefficients that are polynomial in $t$ and rational in $x$ - that the power series $\sum\limits_{n=0}^{\infty}na_nt^n$ satisfies.
Note that
$\sum\limits_{n=0}^{\infty}na_nt^n=\sum\limits_{n=0}^{\infty}\left(nP_n-\left(n-1\right)P_{n-2}\right)t^n=\sum\limits_{n=0}^{\infty}nP_nt^n-\sum\limits_{n=0}^{\infty}\left(n-2\right)P_{n-2}t^n-\sum\limits_{n=0}^{\infty}P_{n-2}t^n$
$=tU^{\prime}-t^3U^{\prime}-t^2U$,
where $U$ denotes the power series $\dfrac{1}{\sqrt{1-2xt+t^2}}=\sum\limits_{n=0}^{\infty}P_nt^n$.
Computations best left to a CAS now yield $\sum\limits_{n=0}^{\infty}na_nt^n=\dfrac{t\left(xt^2+x-2t\right)}{\sqrt{t^2-2tx+1}^3}$. This is the function we wish to find a differential equation for. Now there is a general way for it: If $P$ and $Q$ are two polynomials in $x$ and $t$, then the function $V:=\frac{P}{\sqrt Q}$ satisfies the differential equation $PQ\dfrac{d}{dt}V=\left(P^{\prime}Q - \dfrac{PQ^{\prime}}2\right)V$. This hopefully makes a "nice" recurrence. (In your case $P=t\left(xt^2+x-2t\right)$ and $Q=\left(t^2-2tx+1\right)^3$.)
Best Answer
Nice idea. As far as I'm concerned, the above comments are "answers", since they check out. I might as well record the details: $$ \int_0^1 \frac{dx}{\sqrt{1-2(2x^2-1)t + t^2}} = \frac{1}{2\sqrt{t}} \int_0^1 \frac{dx}{\sqrt{\frac{(1+t)^2}{4t} -x^2}} = \frac{1}{2\sqrt{t}}\arcsin\left(\frac{\scriptstyle 2\sqrt{t}}{\scriptstyle1+t}\right). $$ The half-angle formula for $\sin$ reduces this to $$ \frac{1}{\sqrt{t}} \arcsin \sqrt{\frac{t}{1+t}} = \frac{1}{\sqrt{t}} \arctan \sqrt{t}, $$ which has the desired power series expansion.
Thanks.