A standard way to get a numeric solution is to use the Taylor expansion. Let
$$x^5 + ax + b = 0$$
Be your quintic. Applying the affine transformation $x = \sqrt[4]{a} \cdot z$ gives
$$\sqrt[4]{a^5} \cdot z^5 + \sqrt[4]{a^5} \cdot z + b = 0$$
Dividing out by $\sqrt[4]{a^5}$ and setting $t = b \cdot a^{-5/4}$ result the one-parameter Bring-Jerrard form
$$z^5 + z = -t \tag{+}$$
Write $z = z(-t)$ for one of the 5 solutions of the quintic dependent on parameter $t$. Clearly, if $t = 0$ then $z(t)$ must satisfy $z(z^4 + 1) = 0$ the real solution of which is $z = 0$. Thus, $z(0) = 0$. For calculating $z'(0)$, differentiate $(+)$ with respect to $t$ to get
$$5z(t)^4 \cdot z'(t) + z'(t) + 1 = 0 \Rightarrow z'(t) = \frac1{1 + 5z(t)^4}$$
Substituting $t = 0$ above and noting that $z(0) = 0$, we get $z'(0) = 1$. One can in this way derive the higher derivatives at $t = 0$. First few values of the derivatives are $z(0) = 0, z'(0) = 1, z''(0) = 0, z'''(0) = 0, z^\text{IV}(0) = 0, z^\text{V}(0) = -1$ etc. The nonzero derivatives are $z'(0) = 1, z^\text{V}(0) = -1, z^\text{IX}(0) = 5, z^\text{XIII}(0) = -35$ etc.
Finding a closed form for this is a pain in the neck. Of course,
we can cheat. The first entry is precisely the sequence involved (see Sloan's comment on the formula section). Thus, our desired Maclaurin series is
$$z(t) = \sum_{k = 0}^\infty \binom{5k}{k} \frac{(-1)^{k+1} t^{1+4k}}{1+4k}$$
This is not entirely convergent outside the radius $|t| = 4 \cdot 5^{-5/4}$, so you have to be careful. Let's try out formula for $t = 1/2$ :
Using the
PARI/GP snippet $\colorbox{LightGray}{$\texttt{solve(x=0,-1,x^5+x+1)}$}$ one gets the approximate solution of $-0.47565274353960478548185664$. Whereas using the snippet $\colorbox{LightGray}{$\texttt{sum(k=0,10,binomial(5*k,k)*(-1)^(k+1)*(1/2)^(1+4*k)/(1+4*k)}$}$, one gets $-0.47569803996611881302669644$ which is correct upto 4 decimal digits. This is not really THAT efficient, however one can use some series accelerations to get an rapid convergence of the series.
[PS : We can actually prove the above formula rigorously without any cheating whatsoever, but that'd require a fair bit of complex analysis. Google "Lagrange-Burmann inversion"]
