I’ll answer questions $1$, $3$, $4$ and $5$. The “hyperbolic sinks” in question $2$ are less well-defined than the circular pattern, but if you want to investigate them, the results in this answer should also be helpful for that.
First, there’s a kind of symmetry with respect to reflection in the axial directions. (This was alluded to in the comments, but there were no links to the previous questions, so I’ll derive it here.) We have
\begin{eqnarray}
T^{(n+1)}_{n-i,j}
&=&
\left[x^j\right](1+x)^i(1-x)^{n-i}
\\
&=&
(-1)^j\left[(-x)^j\right](1+x)^i(1-x)^{n-i}
\\
&=&
(-1)^j\left[x^j\right](1-x)^i(1+x)^{n-i}
\\
&=&
(-1)^jT^{(n+1)}_{i,j}
\end{eqnarray}
and
\begin{eqnarray}
T^{(n+1)}_{i,n-j}
&=&
\left[x^{n-j}\right](1+x)^{n-i}(1-x)^i
\\
&=&
\left[x^{-j}\right]x^{-n}(1+x)^{n-i}(1-x)^i
\\
&=&
\left[x^{-j}\right]\left(x^{-1}+1\right)^{n-i}\left(x^{-1}-1\right)^i
\\
&=&
\left[x^j\right](x+1)^{n-i}(x-1)^i
\\
&=&
(-1)^i(x+1)^{n-i}\left(1-x\right)^i
\\
&=&
(-1)^iT^{(n+1)}_{i,j}\;,
\end{eqnarray}
so reflecting in one axial direction incurs a sign alternation according to the coordinate in the other direction.
This reduces questions $4$ and $5$ to question $3$, so what’s left to explain is why in the upper left quarter the values are positive outside the circle and change signs inside the circle.
Roughly, this is because for small $i$ and $j$, $\binom{n-k}j\approx\binom{n-i}j$, independent of $k$, and then the sum is just a binomial expansion:
\begin{eqnarray}
T^{(n+1)}_{i,j}
&=&
(-1)^i\sum_{k=0}^i\binom ik\binom{n-k}j(-2)^k
\\
&\approx&
(-1)^i\sum_{k=0}^i\binom ik\binom{n-i}j(-2)^k
\\
&=&
(-1)^i\binom{n-i}j\sum_{k=0}^i\binom ik(-2)^k
\\
&=&
(-1)^i\binom{n-i}j(-2+1)^i
\\
&=&
\binom{n-i}j\;.
\\
\end{eqnarray}
To try to estimate the area where this keeps the sums positive, we can write
\begin{eqnarray}
\binom{n-k}j
&=&
\binom{n-i}j\cdot\frac{(n-i+1)\cdots(n-k)}{(n-i+1-j)\cdots(n-k-j)}
\\
&=&
\binom{n-i}j\cdot\frac{\left(1-\frac{i-1}n\right)\cdots\left(1-\frac kn\right)}{\left(1-\frac{i-1+j}n\right)\cdots\left(1-\frac{k+j}n\right)}
\\
&=&
\binom{n-i}j\left(1-\frac{(i-k)j}n\right)+O\left(\frac1{n^2}\right)\;.
\end{eqnarray}
Performing the sum over this approximation yields
\begin{eqnarray}
T^{(n+1)}_{i,j}
&=&
(-1)^i\sum_{k=0}^i\binom ik\binom{n-k}j(-2)^k
\\
&\approx&
(-1)^i\sum_{k=0}^i\binom ik\binom{n-i}j\left(1+\frac{(i-k)j}n\right)(-2)^k
\\
&=&
(-1)^i\binom{n-i}j\left(\left(1+\frac{ij}n\right)\sum_{k=0}^i\binom ik(-2)^k-\frac jn\sum_{k=0}^i\binom ik(-2)^kk\right)
\\
&=&
(-1)^i\binom{n-i}j\left(\left(1+\frac{ij}n\right)(-1)^i-\frac jn(-1)^i\cdot2i\right)
\\
&=&
\binom{n-i}j\left(1-\frac{ij}n\right)\;.
\\
\end{eqnarray}
This approximation correctly predicts that there’s an area of small $i$ and $j$ where the sums are positive, and that this area is symmetric under exchanging $i$ and $j$, but the result is qualitatively wrong in that it scales incorrectly – the hyperbola $ij=n$ recedes into the corner as $n$ increases, whereas the circle in the diagrams survives the limit $n\to\infty$ and only depends on $\frac in$ and $\frac jn$.
These considerations suggest that the terms with $(-2)^k$ are unnecessarily oscillating, and things might improve if we resum to a less oscillating representation:
\begin{eqnarray}
T^{(n+1)}_{i,j}
&=&
(-1)^i\sum_{k=0}^i\binom ik\binom{n-k}j(-2)^k
\\
&=&
(-1)^i\sum_{k=0}^i\binom ik\binom{n-k}j\sum_{\ell=0}^k\binom k\ell(-1)^\ell(-1)^{k-\ell}
\\
&=&
(-1)^i\sum_{\ell=0}^i\sum_{k=\ell}^i\binom ik\binom k\ell\binom{n-k}j(-1)^k\;.
\end{eqnarray}
Here we can take a combinatorial approach. As discussed at Generalised inclusion-exclusion principle,
$$
\sum_{k=\ell}^i(-1)^{k-\ell}\binom ik\binom k\ell b_k
$$
is the number of ways to fulfil exactly $\ell$ out of $i$ conditions if there are $b_k$ ways to fulfil $k$ particular conditions. Say we want to choose $j$ out of $n$ objects, but we mustn’t choose $i$ particular objects. Those are $i$ conditions, and there are $\binom{n-k}j$ ways to fulfil $k$ particular ones of them. But of course we can also perform this count directly: To fulfil exactly $\ell$ of the conditions, we can choose $i-\ell$ of the $i$ forbidden objects and the remaining $j-(i-\ell)$ objects from the remaining $n-i$ objects. Thus
$$
\sum_{k=\ell}^i(-1)^{k-\ell}\binom ik\binom k\ell\binom{n-k}j=\binom i{i-\ell}\binom{n-i}{j-(i-\ell)}\;,
$$
so
\begin{eqnarray}
T^{(n+1)}_{i,j}
&=&
\sum_{\ell=0}^i(-1)^{i-\ell}\binom i{i-\ell}\binom{n-i}{j-(i-\ell)}
\\
&=&
\sum_{m=0}^i(-1)^m\binom im\binom{n-i}{j-m}\;.
\end{eqnarray}
With hindsight, this could have been obtained more easily by noting that to get $j$ powers of $x$ from $(1+x)^{n-i}(1-x)^i$ we can choose $m$ from $i$ negative factors and $j-m$ from $n-i$ positive factors, so I’ve merely rederived that representation.
Some rearrangement of the factors brings this into the more useful form
$$
T^{(n+1)}_{i,j}=\binom{n-i}j\sum_{m=0}^\infty(-1)^m\frac{i^{\underline m}j^{\underline m}}{m!}\frac1{(n-i-j+1)^{\overline m}}\;,
$$
where $x^{\overline m}$ and $x^{\underline m}$ are the rising and falling factorials, respectively, and the sum can be extended to infinity because it terminates at ${\min(i,j)}$ due to the falling factorials. This form is only valid for $n-i-j\ge0$, or $i+j\le n$, which holds in the upper left quadrant. By the way, this shows that the signs are symmetric under exchange of $i$ and $j$.
This representation almost has the form of a hypergeometric series – we just have to use the relationship $x^{\underline m}=(-1)^m(-x)^{\overline m}$ to transform the falling factorials into rising ones to obtain
\begin{eqnarray}
T^{(n+1)}_{i,j}
&=&
\binom{n-i}j\sum_{m=0}^\infty(-1)^m\frac{(-i)^{\overline m}(-j)^{\overline m}}{m!}\frac1{(n-i-j+1)^{\overline m}}
\\
&=&
\binom{n-i}j{}_2F_1\left({{-i,-j}\atop{n-i-j+1}};-1\right)\;.
\end{eqnarray}
Now we can apply humankind’s accumulated knowledge about the hypergeometric series. To derive the circle shape in the limit, let $\alpha=\frac in$ and $\beta=\frac jn$ and consider the limit
$$
\lim_{n\to\infty}{}_2F_1\left({{-\alpha n,-\beta n}\atop{(1-\alpha-\beta)n+1}};-1\right)
$$
that determines the sign at the point $(\alpha,\beta)$ in the rescaled diagrams. Such limits can be determined using the method of steepest descent, as shown at Asymptotic behavior of the hypergeometric function and very systematically and exhaustively carried out by R. B. Paris in Asymptotics of the Gauss hypergeometric function with large parameters (Journal of Classical Analysis, vol. $3$, no. $1$ ($2013$), pp. $1$–$15$). The paper is in two parts (available online: I, II); our type of sum is treated in part II.
To apply the paper’s results, we need to first transform the hypergeometric sum using Euler’s transform:
$${}_2F_1\left({{-\alpha n,-\beta n}\atop{(1-\alpha-\beta)n+1}};-1\right)=2^{n+1}{}_2F_1\left({{(1-\beta)n+1,(1-\alpha)n+1}\atop{(1-\alpha-\beta)n+1}};-1\right)\;.$$
The factor $2^{n+1}$ is just another positive factor which, like $\binom{n-i}j$ above, doesn’t affect the signs.
To get to the form treated in the paper, write $\lambda=(1-\alpha-\beta)n$, $\epsilon_1=\frac{1-\beta}{1-\alpha-\beta}$ $\epsilon_2=\frac{1-\alpha}{1-\alpha-\beta}$ and $a=b=c=1$, and without loss of generality assume $\alpha\le\beta$ and thus $\epsilon_1\le\epsilon_2$. That gets us to the case $\epsilon_2\ge\epsilon_1\gt1$ with real argument $z=x$ treated in Section $3.1$ starting on p. $10$ of part II. This has four subcases. Case (i) with $x\gt0$ doesn’t occur here, since we have $x=-1$. The remaining three cases distinguish how $x$ lies with respect to the two boundaries
$$z_*^{\pm}=\frac{\epsilon_1+\epsilon_2-2\epsilon_1\epsilon_2\pm2\sqrt{\epsilon_1\epsilon_2(1-\epsilon_1)(1-\epsilon_2)}}{(\epsilon_1-\epsilon_2)^2}$$
(which are defined in Equation $(2.5)$ on p. $3$). Substituting our $\epsilon_1$ and $\epsilon_2$ yields
$$
z_*^{\pm}=\frac{\alpha(\alpha-1)+\beta(\beta-1)\pm2\sqrt{\alpha(\alpha-1)\beta(\beta-1)}}{(\alpha-\beta)^2}\;,
$$
and substituting the circle $\alpha=\frac12(1-\cos\phi)$, $\beta=\frac12(1-\sin\phi)$ (with $0\le\phi\le\frac\pi2$) into $z_*^+$ shows that it is precisely the curve where $z_*^+=-1$, the boundary between cases (ii) and (iii). The corresponding asymptotic expansions are given in Equations $(3.3)$ on p. $10$ and $(3.4)$ on p. $11$, respectively.
The asymptotic expansion $(3.3)$ for case (ii), which obtains outside the circle, is a single sum. The saddle points
$$
t_{sj}=\frac{2\epsilon_1}{\Delta\pm\sqrt{\Delta^2+4\epsilon_1(1-\epsilon_2)}}
$$
with $\Delta=1+\epsilon_2-\epsilon_1$ (defined in Equation $(2.2)$ on p. $3$) lie in $[0,1]$, so substituting them into Equation $(3.2)$ on p. $10$ yields logarithms of positive real numbers and thus no imaginary part in the phase factor $\psi(t_{s1})$ of the exponential. The remaining coefficients in the leading term are positive real numbers, so in this region the leading term is positive.
By contrast, the asymptotic expansion $(3.4)$ for case (iii), which obtains inside the circle, is a sum of two sums, and the saddle points $t_{sj}$ are complex conjugates, which leads to imaginary parts in the phase factors of the exponentials.
Since the region $i+j\le n$ with $i\le j$ also includes part of the upper right corner, the question arises whether we can recover the alternating signs derived by symmetry above from the asymptotic expansion of the hypergeometric sum, and that’s indeed possible. In this region, case (iv) with the asymptotic expansions $(3.5)$ on p. $11$ and $(3.6)$ on p. $12$ obtains. It’s tempting to see the factor $\sin\pi\nu$ in $(3.5)$ as responsible for the sign alternations, but actually that expansion doesn’t apply here because in our case $\nu$ is a positive integer. Rather, the sign alternations occur because in this region the saddle points lie in $[1,\infty]$, and substituting them into $(3.2)$ leads to the logarithm of a negative number and thus to an alternating sign of the leading exponential term.
