The answer is positive if we add a bit more symmetry in the assuptions (and if I made no mistake !). There is a beautiful theorem by Schlichting :

Theorem (Schlichting). Let $G$ be a group and $\mathfrak H$ a family of subgroups of $G$. Assume that the index $H/H\cap K$ remains bounded for any members $H$ and $K$ of $\mathfrak H$. Then there is a subgroup $N$ of $G$ invariant under the group of automorphisms of $G$ fixing $\mathfrak{H}$ setwise such that $N/N\cap H$ and $H/ N\cap H$ remain bounded for any $H$ in $\mathfrak{H}$.

In every proof I know of Schlichting's Theorem (there are 3 or 4), the subgroup $N$ is obtained as a finite extension of a finite intersection of member of $\mathfrak H$. One can actually do better :

Claim 1. The subgroup $N$ obtained in Schlichting's Theorem is the intersection of finitely many members of $\mathfrak H$.

Corollary 1. $G$ is a group, $H_1,\dots,H_n$ are subgroups of $G$, and $H$ is a subgroup of every $H_i$ such that $H_i/H$ is finite. If every $H_i$ normalises $\bigcap_{i=1}^n H_i$, then $H$ has a subgroup of finite index wich is normal in every $H_i$.

Proof of Corollary 1. Let $\mathfrak H$ the set of $\langle H_1,\dots, H_n\rangle$-conjugates of $H$. By Schlichting's theorem applied to the family $\mathfrak H$ inside the group $I=\bigcap_{i=1}^n H_i$, there is a subgroup $N$ of finite index in $I$ which is normal in every $H_i$. By Claim 1, $N$ is a finite intersection of $\langle H_1,\dots, H_n\rangle$-conjugates of $H$ so it must be a subgroup of $H$.

Question. Is Corollary 1 a trivial statement ?

A compactness argument -using the axiom of choice !!- can show that the index of $N$ can be bounded by a constant depending only on the index of $H$ in $\bigcap_{i=1}^n H_i$ (that is for sure too strong a method, but I do not know how to find this optimal bound).

Let us proove Claim 1 now.

Definition 1 (Wagner). Let $\mathcal L$ be a lattice. A rank on $\mathcal L$ is a function from $\mathcal L^2$ to $\mathbf N\cup\{\infty\}$ satisfying the following properties:

1. There is some $k$ in $\mathbf N$ such that $\delta(a,a)$ equals $k$ for all $a$ in $\mathcal L$.

2. $\delta$ is increasing in the first argument and decreasing in the second.

3. If $a'\geq a\geq a\geq b\geq b'$ are elements of $\mathcal L$, if $\delta(a,b)$ and $\delta(a',b')$ are equal and finite, then $a=a'$ and $b=b'$.

4. There is an increasing function $g$ from $\mathbf{N}^2$ to $\mathbf N$ such that whenever $a\geq b\geq c$ and both $\delta(a,b)$ and $\delta(b,c)$ are finite, then $\delta(a,c)\leq g(\delta(a,b),\delta(b,c))$.

5. There is an increasing function $f$ from $\mathbf N^2$ to $\mathbf N$ such that whenever both $\delta(a,c)$ and $\delta(b,c)$ are finite, then $\delta(a\lor b,c)\leq f(\delta(a,c),\delta(b,c))$.

Theorem 1 (Wagner). $\mathcal L$ is a lattice with rank $\delta$ and $\mathfrak F$ is a family of elements in $\mathcal L$ such that $\delta(a_0,a_1\land\dots\land a_i)\leq n(i)$ for all $i$ in $\mathbf N$ and $a_0,\dots,a_i$ in $\mathfrak F$. There is some $f$ in $\mathfrak F$ and $m$ in $\mathbf N$ such that $f$ is fixed by all automorphisms of $\mathcal L$ fixing $\mathfrak F$ setwise and leaving $\delta$ invariant, with in addition $\delta(a,f)\leq m$ and $\delta(f,a)\leq f(n(1),f(k,n(1)))$. More precisely, $f$ is a finite intersection $\bigwedge_{i=1}^ma_i\lor a_{m+1}$ where $a_0,\dots,a_m,a_{m+1}$ are finite intersections of members of $\mathfrak F$.

Using Mark Sapir's idea here, one can say:

Lemma 1. $G$ is a group, $H$ and $K$ are subgroups of $G$ and $\sigma$ is an group automorphism of $G$. If $\sigma$ stabilises $H\cup K$ setwise, then either $H\cup K$ is a group or $\sigma$ stabilises $H\cap K$.

Proof of Lemma 1. The group $\sigma H$ is the union of the two groups $\sigma H\cap H$ and $\sigma H\cap K$. This can happen only if either $\sigma H=\sigma H\cap H$ or $\sigma H=\sigma H\cap K$. In the first case, $\sigma H\subset H$, and in the second $\sigma H\subset K$. Similarly, either $\sigma K\subset K$, or $\sigma K\subset H$. They are four cases to deal with and in fact two by symmetry. Assume first that $\sigma K\subset H$ and $\sigma H\subset H$. Then $K\subset\sigma^{-1}H$ and $H\subset\sigma^{-1}H$. As $\sigma^{-1}$ stabilises $H\cup K$, we either have $\sigma^{-1}H\subset H$ or $\sigma^{-1}H\subset K$, so the first case leads to $K\subset H$ or $H\subset K$. Second case : $\sigma K\subset H$ and $\sigma H\subset K$ which yield $\sigma (H\cap K)\subset (H\cap K)$. This also holds for $\sigma^{-1}$, so $\sigma(H\cap K)=H\cap K$.

Proof of claim 1. We call a coset in $G$ a left coset $gH$ of any subgroup $H$ of $G$ and consider the set $\mathcal C$ of finite unions of cosets in $G$. By convention, an empty union is empty so the emptyset is in $\mathcal C$. The set $\mathcal C$ is partially ordered by inclusion. If $gH$ and $gK$ are two cosets in $G$, the intersection $gH\cap gK$ is either empty or a coset of $H\cap K$. Equiped with intersection and union, $\mathcal C$ forms a distributive lattice.

For any $A$ in $\mathcal C$, we say that a subgroup $H$ of $G$ is represented in $A$ if $gH\subset A$ for some $g$ in $G$. If $B\subset A$ and $B$ is non empty, the group $\{1\}$ is represented in $B$, so it is always possible to find a familly $(H_i)_{i\in I}$ of subgroup represented in $B$ and elements $(g_i)_{i\in I}$ in $G$ such that$\bigcup_{i\in I} g_i H_i\cup B=A$. We call such a union a $B$-covering of $A$ of size $|I|+1$. If $B\subset A$, we define the rank $\delta(A,B)$ to be the minimal size of $B$-coverings of $A$. For arbitrary $A$ and $B$ in $\mathcal C$, we extend the definition by putting $\delta(A,B)$ equal to $\delta(A,A\cap B)$. Note that every automorphism of $G$ leaves $\delta$ invariant. If $H$ and $K$ are two subgroups of $G$, then $\delta(H,K)$ is the usual index $[H:H\cap K]$. It is not difficult to check that this is indeed a rank in the sense of Definition 1, taking addition and multiplication for $f$ and $g$.

Now consider $\mathcal L$ the sub-lattice of $\mathcal C$ generated by the elements in $\mathfrak H$. The number $\delta(H,H')$ is bounded by $n$ for every $H,H'$ in $\mathfrak H$. By Theorem 1, there is some $L$ in $\mathcal L$ and a natural number $p$ such that $\delta(H,L)$ is at most $p$ and $\delta(L,H)$ is at most $2n+1$ for all $H$ in $\mathfrak H$, and such that $L$ is invariant by any automorphism of $G$ fixing $\mathfrak H$ setwise. But, as the lattice is distributive, $L$ is also the union of two groups $A$ and $B$ commensurable with $\mathfrak H$. By Lemma 1 either $A\cup B$ or $A\cap B$ do the job.The End

Corollary 2. Your question has a positive answer if and only if there is at least one subgroup $L$ of $G$ commensurable with $H_1$ and normalised by every $H_i$.

Proof. If $K$ is of finite index in every $H_i$, set $I=L\cap K$ and apply Schlichting to the set of $\langle H_1,\dots, H_n\rangle$-conjugates of $I$ inside $L$.