Write $X\_N$ for this blow up. Place the N points in 'general position' as needed. Then $X\_6$ embeds in $CP^2$ as a smooth cubic surface. (See, eg, Griffiths and Harris.) But there is no other $N$ (except $N=0$)
for which $X\_N$ embeds in $CP^3$.
(Proof: The topology of the blow-up
disagrees with that of a smooth surface of degree $d$ in $CP^3$. (Gompf-Stipsisz p. 21.) On the other hand, $X\_N$ embeds in $CP^5$ simply because any
smooth algebraic surface $X$ so embeds. (Harris, `Algebraic Geometry, a first course', p. 193.) 


Embarrassingly, I don't even know the answer for $N=1$ where $X\_1$ is the 1st Hirzebruch surface! (I'm betting it does embed.)


Motivation: This question began in an attempt to better understand the 27 lines on the cubic
and my initial surprise at how the construction described in GH of $X\_6$ yielded 
a smooth surface in $CP^3$, and how all such surfaces arise through that construction by varying the 6 points. I am hoping answers might help me understand the moduli of blow-ups as I move the N points about the plane, and orient me as a novice to algebraic surfaces. 



When does  the blow-up of $CP^2$ at N points  embed in $CP^4$?
For $N=1$ the answer is yes: the embedding into ${\mathbb P}^4$ is given by the linear system of conics through the blown up point (the image has degree $d=3$).
For $N=5$, the system of cubics through the 5 points gives an embedding ($d=4$).


ADDED: here are 2 slightly less obvious examples:
For $N=8$ one can take quartics with an assigned double point and 7 simple base points ($d=5$).
For $N=10$ take the quintics with 3 assigned double points and 7 simple base points ($d=6$; I did not check all the details here, because it's very boring, but I'm sure that it works).


In general, giving a satisfactory answer to your question seems very hard. There is a numerical equality,
 the so-called "double point formula" (Hartshorne, "Algebraic geometry", p.434), which is satisfied by all smooth surfaces of ${\mathbb P}^4$:
$$d^2-10d-5HK+12\chi-2K^2=0,$$
where $H$ is the hyperplane section, $d=H^2$ is the degree, $K$ is the canonical divisor and $\chi$ the Euler characteristic of ${\mathcal O}\_{X\_N}$.
In our case the formula becomes:
$$d^2-10d-5HK+2N-6=0.$$
In addition there is result by G. Ellingsrud and C. Peskine [Invent. Math. 95 (1989), no. 1, 1--11] saying that only finitely many components of the Hilbert scheme of smooth surfaces in ${\mathbb P}^4$ contain smooth rational surfaces.
 So in principle it should be possible to classify all the smooth rational surfaces in ${\mathbb P}^4$. In practice, it is known that the degree is $\le 76$, 
[Cook, An improved bound for the degree of smooth surfaces in P4 not of general type.
Compositio Math. 102 (1996)] and it is conjectured that $d\le 15$ (examples with $d=15$ do exist). There are also papers by several authors (Ranestad, Schreyer, Popescu, and others) that classify the smooth rational surfaces of ${\mathbb P}^4$ of degree $\le 11$. In these papers you can find examples of the kind you are looking for. For instance there are examples with $d=10$ and $N=18$.


