This is an old question but I hope the following is still of interest.

If $M^n$ is a closed simply connected manifold then the inclusion $M^n\backslash \{ pt\}\hookrightarrow M^n$ is $(n-1)$-connected which means that the induced map of minimal models is an isomorphism through dimension $(n-2)$. Next note that $M^n\backslash \{ pt\}$ has zero homology in degrees above $n-2$. It's a general fact that given a minimal model up to dimension $k$ of a space whose cohomology vanishes in degrees above $k$ the rest of the minimal model is determined uniquely (and constructively) from the model up to degree $k$. This provides an easy recipe for computing the minimal model of
$M^n\backslash \{ pt\}$ which can be more explicitly described as follows.

If $(\Lambda V, d)$ is a minimal model of $M^n$ then consider the following dga $(A,d)=(\Lambda V\oplus\Lambda \langle z\rangle/(z^2), d)$ with $\deg z=n-1, V\cdot z=0$ and $dz=[M]$ - the fundamental class of $M$. This is a model (non-minimal and even a non-free one!) of $M^n\backslash \{ pt\}$. In practice it's easier to directly compute the minimal model of $A$ by the general procedure outlined above.

Here are a couple of examples.

Let $M^4=\mathbb {CP}^2$. Its minimal model is $(\Lambda \langle x,y\rangle,d)$ with deg x=2, deg y=5, dx=0, dy=x^3. Up to degree $n-2=2$ this is simply given by $\Lambda\langle x\rangle$ with dx=0. Next we need another generator to make $H^4=0$ (which is currently generated by $[M]=x^2$) so we add $z$ of deg $3$ such that $dz=x^2$. Now the model $(\Lambda \langle x,z\rangle,d)$ with deg x=2, deg z=3, dx=0, dy=x^2 already has $H^i=0$ for $i\ge 4$ so we don't need to add anything else. The resulting model is easily recognized as the model of $\mathbb S^2$ which is of course not surprising since $\mathbb{CP}^2\backslash\{pt\}$ is a Hopf disk bundle over $\mathbb{CP}^1$.

A more interesting example: Let $M=\mathbb S^3\times\mathbb S^5$. Its minimal model is generated by $x,y$ with $\deg x=3, \deg y=5$ and $dx=0,dy=0$. Applying our recipe the model of $\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$ will be the same through dimension 6. Next,
we need to kill off cohomology in degree 8 which is currently generated by $[M]=xy$. So we need another generator $z$ of degree 7 with $dz=xy$. However, adding such generator introduces more cohomology in degrees 10 and 12 generated by $xz$ and $yz$. So we need two more generators $a$ and $b$ with $da=xz, db=yz$. However, adding those introduces yet more cohomology and we need to keep adding more generators. This will continue forever because $\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$ is rationally hyperbolic.

Lastly, let me mention that operations such as cell attachments (or in this case cell deletions) are usually easier handled by Quillen Lie algebra models which are better suited to work with cofibrations (while Sullivan models are better suited for fibrations). In this particular case it's especially easy. If $(\mathbb L_V,d)$ is a minimal Quillen Lie model of $M^n$ then the model of $M^n\backslash \{ pt\}$ is obtained by simply removing a single generator from $V$ corresponding to the fundamental class of $M$.