Any symplectic linear transformations in $T_xM$ is locally realizable as a Hamiltonian vector field, thus for questions 1 and 2, one can profitably use representation theory of the symplectic group.
FACT (Lefschetz decomposition) Let $W$ be a $2n$-dimensional symplectic vector space, $\bigwedge^\ast W$ its exterior algebra, and $\omega\in\bigwedge^2 W$ the invariant two-form. Exterior multiplication by $\omega$ and the contraction with $\omega$ define a pair of $Sp(W)$-equivariant graded linear transformations $L, \Lambda$ of $\bigwedge^\ast W$ into itself of degrees $2$ and $-2,$ and let $H=\deg-n$ be the graded degree $0$ map acting on $\bigwedge^k$ as multiplication by $k-n.$ Then $L,H,\Lambda$ form the standard basis of the Lie algebra $\mathfrak{sl_2}$ acting on $\bigwedge^\ast W$ and the actions of $Sp(W)$ and $\mathfrak{sl_2}$ are the commutants of each other.
See, for example, Roger Howe, Remarks on classical invariant theory.
Corollary Every homogeneous $Sp(W)$-invariant element of $\bigwedge^\ast W$ is a multiple of $\omega^k$ for some $0\leq k\leq n.$
Since, conversely, every polynomial in $\omega$ is invariant under the Hamiltonian vector fields, this gives a full description of the invariant differential forms.
For question 2, locally every invariant tensor must reduce to an $Sp(W)$-invariant element of the tensor algebra. For the special case of symmetric tensors, the answer is trivial.
FACT Under the same assumptions, the $k$th symmetric power $S^k W$ is a simple $Sp(W)$-module (non-trivial for $k>0$).
General case can be handled using similar considerations from classical invariant theory. A more involved question of describing the invariant local tensor operations on symplectic manifolds (an analogue of the well-known problem of invariant local operations on smooth manifolds, such as the exterior differential or Schoutens bracket) was considered in an old article by A.A.Kirillov.
1) No. There are many more vector fields. The vector fields you are looking for are precisely those which vanish at $p_0$. Since $h^0( \mathbb P^n, T \mathbb P^n) = (n+1)^2 -1$ and you are imposing $n$ linearly independent conditions you should get $n^2 + n - 1$ vector fields. In homegeneous coordinates they can be written as
$$
l_0 \frac{\partial}{\partial z_0} + \sum_{i=1}^n \sum_{j=1}^n c_{ij} z_i \frac{\partial}{\partial z_j} .
$$
Notice that the homogeneous vector field $\sum_{i=0}^n z_i \frac{\partial}{\partial z_i}$ does not contribute to the counting since it corresponds to the zero vector field on $\mathbb P^n$.
2) You should get $k^2 + (n+1-k) n -1$ vector fields. As pointed out in the comments the
result remains unchanged if we swap $k$ and $n-k+1$. The point is that $PGL(n+1)$ acts on $\mathbb P^n$ as well as on the dual projective space $\check{\mathbb P}^n$. If we look at the subgroup preserving a linear subspace of codimension $k$ on $\mathbb P^n$, then the natural action on $\check{\mathbb P}^n$ will preserve its dual: a linear subspace of codimension $n-k+1$.
3) You will get a holomorphic vector field on the blow-up if and only if the curve is left invariant by the original vector field.
If your vector field vanishes on a linear subspace of dimension at least two then every curve contained in it will be invariant by the vector field. Thus on the blow-up along these curves you will still get holomorphic vector fields.
If instead you restrict your attention to vector fields with zero set of dimension at most one then while you can find curves of arbitrarily high degree invariant by this class of vector fields (think on orbits of $\mathbb C^*$-actions on $\mathbb P^n$), they all have (geometric) genus zero.
Best Answer
I guess there will not exist a $\tilde \varphi_X \in C^{\infty}(\tilde M, \mathbb{R})$ if $\omega_\epsilon$ is not invariant under $Re(X)$. But if you allow $C^{\infty}(\tilde M, \mathbb{C})$, then the following general fact is true :
$\textbf{Proposition}:$ A holomorphic vector field has a hamiltonian (possibly complex valued) (with respect to any Kahler form) if and only if it vanishes at some point.
This fact is an exercise in "An introduction to extremal Kahler metrics" by Gabor Szekelyhidi. He attributes this fact to LeBrun-Simanca.
One possible way of proving the Proposition is the following - Let $(M,\omega)$ be a Kahler manifold and $X$ a holomorphic vector field such that $X(p)=0$ for some $p\in M$. Since $X$ is holomorphic, $i_X\omega$ is $\bar\partial$-closed $(0,1)$ form, and represents a co-homology class. So there must exist a unique harmonic $(0,1)$ form $\alpha \in [i_X\omega]$. If we can show that $\alpha = 0$, we are done, since then $i_X\omega$ would have to be $\bar \partial$-exact which is equivalent to saying that $X$ is Hamiltonian.
$\textbf{Claim}$ : $\bar \alpha (X) = 0$. To see this, we first notice that since we are working on Kahler manifold and $\alpha$ is $\bar\partial$-harmonic $(0,1)$ form, $\bar\alpha$ is a $\bar\partial$-harmonic $(1,0)$ form. So $\bar\partial \bar\alpha = 0$. But then this implies that $\bar\alpha(X)$ is a holomorphic function which vanishes at $p\in M$, and so must be identically zero. We can now complete the proof -
$$\int_{M}|\alpha|^2\frac{\omega^n}{n!} = \int_{M}\bar\alpha\wedge\alpha\wedge\frac{\omega^{n-1}}{(n-1)!} = \int_{M}\bar\alpha \wedge i_{X}\omega \wedge \frac{\omega^{n-1}}{(n-1)!} = \int_{M}\bar\alpha(X)\frac{\omega^n}{n!}=0.$$
The second equality follows from integration by parts, since $\alpha = i_X\omega + \bar\partial f$ for some function $f$, and $\omega$ and $\bar\alpha$ are $\bar\partial$-closed. Hence $\alpha = 0$, and this completes the proof of the Proposition.
So in your problem since the lifted vector does vanish (on all exceptional divisors), it will have a (possibly complex valued) hamiltonian with respect to any Kahler form on $\tilde M$.