Explicit equations for $Y(N)$ for small $N$ Consider the congruence subgroup
$$\Gamma(N) = \left\{\left(\begin{array}{cc}
                            a & b \\
                             c & d\end{array} \right) \in SL_2(\mathbb{Z})\ ;\ \left(\begin{array}{cc}
                            a & b \\
                             c & d\end{array} \right) \equiv  \left(\begin{array}{cc}
                            1 & 0 \\
                             0 & 1\end{array} \right)\mod N\right\}$$
acting on the upper half-plane $\mathfrak{h} = \{\tau \in \mathbb{C} \ ;\  \Im(\tau)>0\}$ by fractional linear transformations:
$$ 
\left(\begin{array}{cc}
                            a & b \\
                             c & d\end{array} \right) \cdot \tau = \frac{a\tau + b}{c\tau + d}
$$
It is known that the quotient $Y(N) = \Gamma(N)\backslash\mathfrak{h}$ is an algebraic curve for every $N$. For instance,
$$Y(2) = \mathbb{P}^1 \setminus \{0,1,\infty\}$$
which can be proved by considering modular function $\lambda : \mathfrak{h} \to \mathbb{C}$. It is also well known that the compactification $X(7)$ of $Y(7)$ is isomorphic to Klein's quartic in $\mathbb{P}^2$.
What about the other small values of $N$? How do I compute $Y(N)$ for $N=3,4,5,6$? I know I can look at the genus of the compactification $X(N)$, but is there a way of getting the precise form of $Y(N)$ as in the example for $N=2$?
Bonus: What about the universal elliptic curve over $Y(N)$, is it also computable in these small cases?
 A: Let me try to give a complete answer.
First, for $ N = 3,4,5 $ the genus of the compactification $X(N)$ is $0$, therefore it is isomorphic to $\mathbb{P}^1$. Since $X(N) = Y(N) \cup C(N)$, where $C(N)$ is the finite set of cusps of $\Gamma(N)$, namely $C(N) = \Gamma(N) \backslash \mathbb{P}^1(\mathbb{Q})$, we see that $Y(N) = \mathbb{P}^1 \setminus C(N)$, where $C(N)$ is a finite set of points. Specifically, computing the number of cusps, we see that $|C(2)| = 3, |C(3)| = 4, |C(4)| = 6$ and $|C(5)| = 12$.
Moreover, we know that the isomorphism $X(1) \rightarrow \mathbb{P}^1$ is given by the $j$-map. We can compute the natural covering map $X(N) \rightarrow X(1)$, and by composition obtain the natural corresponding isomorphism. (This is the modular function $\lambda : \mathfrak{h} \rightarrow \mathbb{C}$ you have described above). This map also allows you to represent the universal elliptic curve, as it describes $j$ as a rational function in $\lambda$, and one may simply substitute it in the description of $E_j$.
This was done for several cases, using towers of prime-power level subgroups, basically using two methods.
In the paper https://projecteuclid.org/euclid.ant/1513090725 [Sutherland, Andrew; Zywina, David. Modular curves of prime-power level with infinitely many rational points. Algebra Number Theory 11 (2017), no. 5, 1199--1229. doi:10.2140/ant.2017.11.1199] the authors used Siegel functions to describe the $j$-map for $X(2), X(3), X(4), X(5)$. The tables in the end of the paper show the covering maps to intermediate covers, but composing them all together one obtains the result.
In the paper https://link.springer.com/article/10.1007/s40993-015-0013-7 [Rouse, J., Zureick-Brown, D. Elliptic curves over ℚ
and 2-adic images of Galois. Res. Number Theory 1, 12 (2015). https://doi.org/10.1007/s40993-015-0013-7] the authors used Eisenstein series to describe the $j$-map for $X(2)$ and $X(4)$, again by maps to intermediate covers.
Finally, for $N = 6$, the genus of $X(6)$ is $1$, and since it has rational points, we know it is an elliptic curve. As $\Gamma(6)$ has $12$ cusps, $Y(6)$ is this elliptic curve without $12$ points.
Generalizing and adapting the methods of Rouse and Zureick-Brown, I calculated that $X(6)$ is the Elliptic Curve $y^2 = x^3 + 1$, and the cusps are mapped to the $12$ points
$$ S = \left \{ \infty, (-1,0), (2, \pm 3), (0,\pm 1), (\zeta_6^{\pm1},0
), \left(-2 + \sqrt{-3}, \pm (3 + \sqrt{-3}) \right), \left(-2 - \sqrt{-3}, \pm (-3 + \sqrt{-3}) \right) 
\right \}$$ so that
$$
Y(6) = E \setminus S
$$
I also ran it for the other curves above to see where the cusps map to. One obtains the following models.
$$
Y(3) = \mathbb{P}^1 \setminus \left \{ 0, \frac{1}{3}, \frac{1}{2} \left( 1 \pm \frac{\sqrt{-3}}{3} \right) \right \}
$$
$$
Y(4) = \mathbb{P}^1 \setminus \left \{ 0, \pm 1, \infty, \pm i \right \}
$$
$$
Y(5) = \mathbb{P}^1 \setminus \left \{ 0, \infty, \phi \zeta_5^j, -\phi^{-1} \zeta_5^j \right \}
$$
where $\phi = \frac{1 + \sqrt{5}}{2}$, and $\zeta_5$ is a $5$-th root of unity.
As for the universal elliptic curve, it's slightly tedious to explicitly write it down here, but it is simply substituting the $j$-map in each of the cases.
